Natural numbers as a subset of integer numbers: $\mathbb{N}\subset\mathbb{Z}$. Within set theory, having the natural numbers $\mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $\mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $\mathbb{N}\times\mathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.
Is there any alternative definition of the set $\mathbb{Z}$, starting from $\mathbb{N}$ already defined as usual, such that $$\mathbb{N}\subset\mathbb{Z}$$
as sets, preserving the sum and product operations?
 A: Yes, one can construct a model of $\mathbb{Z}$ which contains $\mathbb{N}$. 
The following construction does not use equivalence classes or embedding technic to make $\mathbb{N}$ subset $\mathbb{Z}$. It rather extends a particular model of natural numbers.
Introduction. Let $(\mathbb{N},\mathrel{+} ,\cdot , \leq)$ be the system of Von Neumann natural numbers. 
A natural number $m$ is a particular set with $m$ elements $m = \{0,...,m-1\}$ for $m>0$ and $0=\{\}$. 
Subtraction and division can be defined for some pairs of natural numbers. 
For $m,n \in \mathbb{N}, m - n$ is the natural number $d$, if there is any, such that $m = n + d.$
For $m,n \in \mathbb{N}, n>0, m\div n$ is the natural number $q$, if there is any, such that $m=n⋅q$.
In what follows I will use the fact that $n\subset\mathbb{N}$ and carry set-theoretical operations on natural numbers.
Construction of Integers
Definition 1. Let $n\in \mathbb{N}$ be a natural number. An opposite number $\overline{n}$ is a subset of $\mathbb{N}$ defined as:
$$\overline{n}:=\begin{cases}0&\text{if } n=0\\ \mathbb{N}\setminus n &\text{if } n\neq 0.\end{cases}$$
The set of all opposite numbers is denoted by $\overline{\mathbb{N}}=\{\overline{n}|n\in\mathbb{N}\}$.
Intuition 1.
An opposite number $\overline{n}$ is a particular set with $n$ elements being missing. Intuitively if we are missing $n$ elements and we receive $n$ then we do not miss anything and therefore we have nothing. This justifies our definition of $\overline{0}=0$.
Definition 2.
We define define the set $\mathbb{Z}$ of integers as
$ \mathbb{Z}=\mathbb{N}\cup \overline{\mathbb{N}}$.
We extend the domain of our definition of $\overline{a}:=\mathbb{N}\setminus a$ to all $a\in\overline{\mathbb{N}}\setminus\{0\}$.
Definition 3.
We define projection functions
$$\mathsf{proj}_0: \mathbb{Z} \to \mathbb{N},a \mapsto a_0:=
\begin{cases}
a, & \text{if } a \in\mathbb{N} \\[2ex]
0, & \text{if } a \in\overline{\mathbb{N}}
\end{cases}
$$
$$\mathsf{proj}_1: \mathbb{Z} \to \mathbb{N},a \mapsto a_1:=
\begin{cases}
\overline{a}, & \text{if } \overline{a} \in\mathbb{N} \\[2ex]
0, & \text{if } \overline{a} \in\overline{\mathbb{N}}
\end{cases}
$$
Definition 4. We define the balance function as follows
$$
\mathsf{bal}: \mathbb{N}\times\overline{\mathbb{N}}\to\mathbb{Z}, (m,\overline{n})\mapsto (m-\mathrm{min}\{m,n\})\cup(\overline{n-\mathrm{min}\{m,n\}}).
$$
The balance function is well-defined as either $m-\mathrm{min}\{m,n\}=0$ or $\overline{n-\mathrm{min}\{m,n\}}=0$.
Intuition 2.
For a natural number $m$ and an opposite number $\overline{n}$ we find a balance between $m$ and $\overline{n}$. If we are missing $n$ elements and we receive $m$ then we have $m-n$ elements if $m<n$, we don’t have or don’t miss any elements if $m=n$ and finally we miss $n-m$ elements if $m>n$.
Definition 5.
We define,$+_\mathbb{Z}, \cdot_\mathbb{Z}$binary operations and $\leq_\mathbb{Z}$ an order on $\mathbb{Z}$ as follows:
$$+_\mathbb{Z} :\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}, (a,b)\mapsto \mathsf{bal}(a_0+b_0,\overline{a_1+b_1})$$
$$\cdot_\mathbb{Z}:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z},  (a,b)↦(a_0\cdot b_0+ a_1\cdot b_1 )\cup(\overline{ a_0\cdot b_1+ a_1\cdot b_0}) $$
$$ a \leq_\mathbb{Z} b :\Longleftrightarrow a_0 + b_1\leq a_1 + b_0.$$
Again as either $a_0\cdot b_0 + a_1\cdot b_1=0$ or $\overline{a_0\cdot b_1+ a_1\cdot b_0}=0$ and the binary operation $\cdot_\mathbb{Z}$ is well-defined.
Proposition 3.
The binary operations and the order on $\mathbb{Z}$ restricted to natural numbers are the same as the binary operations and the order on $\mathbb{N}$.
As a bonus I present a construction of the rational numbers in the same spirit.
Construction of Rationals
Definition 6. Let $m, n\in \mathbb{N}$ and $n>0$. A ratio of $m : n$ is a subset of $\mathbb{N}$ defined as follows: 
$$m:n=(m+n)\div\mathrm{gcd}\{m,n\}\setminus\{m \div\mathrm{gcd}\{m,n\}\}.$$
The set of ratios is the set $\mathbb{L}:=\{m:n|m, n \in \mathbb{N} \text{ and } n\neq 0\}.$ (Ancient and Modern λόγος (lógos) ‘ratio’.)
Intuition 3. We represent natural numbers $m,n$ as intervals $[0; m), [0; n)$ then the ratio of $[0; m) : [0; n)$ is the same as ratio $[0; m), [m; m+n)$. A ratio is a partion of $[0,m+n)$ which we represent by removing the point $m$ from $[0; m+n)$, i.e. $[0; m+n)\setminus \{m\}=[0; m)\cup(m; m+n)$.
Proposition 4.
For coprime natural numbers $m, n \in \mathbb{N}$ and $n>0$
$$m : n := m\cup ((m+n)\setminus (m+1)).$$
For all natural numbers $m \in \mathbb{N}$: $$m : 1 = m.$$
We can now define addition, multiplication and the order on $\mathbb{L}$.
Definition 7.
We define 
$$+_\mathbb{L} :\mathbb{L} \times \mathbb{L} \to \mathbb{L}, (a,a')\mapsto (m\cdot n'+m'\cdot n ) : (n\cdot n' )$$
$$\cdot_\mathbb{L} :\mathbb{L}\times \mathbb{L} \to \mathbb{L},  (a,a')\mapsto (m\cdot m'):(n\cdot n' ) $$
$$ a \leq_\mathbb{L} a' :\Longleftrightarrow m\cdot n'\leq m'\cdot n .$$
Definition 8.
Let $a=m:n, a'=m':n'$ be ratios with $a\leq_\mathbb{L}a'$. We define subtraction
$$a'-_\mathbb{L}a:=(m'\cdot n-m\cdot n' ) : (n'\cdot n ). $$
The order condition on the ratios is what is needed for the subtraction of natural numbers to be well-defined.
Proposition 5.
The set of natural numbers is subset of $\mathbb{L}$ and operations of addition, multiplication, subtractions and the linear order on $\mathbb{L}$ extend those on $\mathbb{N}$.
Definition 9. Let $a\in \mathbb{L}$ be a ratio. An opposite ratio $\overline{a}$ is a subset of $\mathbb{N}$ defined as:
$$\overline{a}:=\begin{cases}0&\text{if } a=0\\ \mathbb{N}\setminus n &\text{if } a\neq 0.\end{cases}$$
The set of all opposite ratios is denoted by $\overline{\mathbb{L}}=\{\overline{a}|a\in\mathbb{L}\}$.
Definition 10.
We define the set $\mathbb{Q}$ of rational numbers as
$$ \mathbb{Q}=\mathbb{L}\cup \overline{\mathbb{L}}.$$
We repeat Defintions 3, 4, 5 subsituting $\mathbb{N}$ with $\mathbb{L}$, $\mathbb{Z}$ with $\mathbb{Q}$, and use operations defined on the set of ratios, rather than on the set of natural numbers.
Proposition 6.
$$\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}.$$
The binary operations and the order on $\mathbb{Q}$ restricted to natural numbers are the same as the binary operations and the order on $\mathbb{N}$.
The binary operations and the order on $\mathbb{Q}$ restricted to integers are the same as the binary operations and the order on $\mathbb{Z}$.
Definition 11.
We define the fraction function as follows
$$\mathsf{frac}:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Q},$$
$$  (a,b)↦\frac{a}{b}:=(a_0\cdot b_0+ a_1\cdot b_1 ):(b_0^2+b_1^2)\cup\overline{(a_0\cdot b_1+ a_1\cdot b_0):(b_0^2+b_1^2)} $$
Again as either $a_0\cdot b_0 + a_1\cdot b_1=0$ or $\overline{a_0\cdot b_1+ a_1\cdot b_0}=0$ so the fraction function is well-defined.
I don't know how this construction can be extended any further but there are various unique representations of reals as continuous fraction so I think it's a possibility that we can construct real numbers as subsets of $\mathbb{N}$.
As a bonus 2 I recap below Dedekind constructions on  $\mathbb{L}$. The following variant is due to Holmes, page 95. Elementary Set Theory with a Universal Set http://math.boisestate.edu/~holmes/holmes/head.pdf.
Construction of Reals
Definition 11.
A magnitude $x$ is a proper initial segment of $\mathbb{L}$ with no greatest element.
The set of magnitudes is the set $$\mathbb{M}:=\{x\subset\mathbb{L}| x \neq \mathbb{L}, \text{for all } a\in \mathbb{L}, a\in x \Longleftrightarrow \text{ for some } b\in x: a<b \}.$$
We define addition, multiplication and the order on $\mathbb{M}$.
Definition 12.
We define 
$$+_\mathbb{M} :\mathbb{M} \times \mathbb{M} \to \mathbb{M}, (x,y)\mapsto \{a+ b|a\in x, b\in y\}$$
$$\cdot_\mathbb{M} :\mathbb{M}\times \mathbb{M} \to \mathbb{M},  (x,y)\mapsto \{a\cdot b|a\in x, b\in y\} $$
$$ x \leq_\mathbb{M} y :\Leftrightarrow x\subset y.$$
Definition 13.
Let $x, y$ be magnitudes with $x\leq_\mathbb{L}y$. We define substruction
$$y-_\mathbb{M}x:=\{b-a|b\in y \text{ and } a\notin x\}. $$
Definition 14. Let $x\in \mathbb{M}$ be a magnitude. An opposite magnitude $\overline{x}$ is a subset of $\mathbb{L}$ defined as:
$$\overline{x}:=\begin{cases}0&\text{if } x=0\\ \mathbb{L}\setminus x &\text{if } x\neq 0.\end{cases}$$
The set of all opposite magnitudes is denoted by $\overline{\mathbb{M}}=\{\overline{x}|x\in\mathbb{M}\}$.
Definition 15.
We define the set $\mathbb{R}$ of real numbers as
$$ \mathbb{R}=\mathbb{M}\cup \overline{\mathbb{M}}.$$
From this moment we carry forward the same as for $\mathbb{Z}$ and $\mathbb{Q}$ and again we re-use defintions 3, 4, 5 with analogous changes. 
Question.
We can construct a model of $\mathbb{Z}$, $\mathbb{Q}$ where all integers, rationals are subsets of $\mathbb{N}$. 
Can we construct a model of $\mathbb{R}$ where all reals are subsets $\mathbb{N}$?
Update 1
I've looked at the constructions of real numbers via continued fractions and I think the answer is one can code real numbers as subsets. 
G. J. Rieger. A new approach to the real numbers (motivated by continued fractions). AOh.
Brauceig. Wis. Ge, 33:205–217, 1982.
A. Knopfmacher and J. Knopfmacher. Two constructions of the real numbers via alternating
series. International Journal of Mathematics and Mathematical Sciences, 12(3):603–613,
1989.
Definition 16
Let $a:N\to\mathbb{N}$ be a sequence of natural numbers where $N\in \mathbb{N}$ or $N= \mathbb{N}$ such that $a_{N-1}>1, N\in\mathbb{N}$. 
We define recurslively a sequence $ q:N \to\mathbb{N}$, $q_0=1$, $q_1=a_1$, $q_n=a_n\cdot q_{n-1}+q_{n-2}$ for $ n\geq2$. A continued ratio  is a subset of $\mathbb{N}$ defined as follows
$$ \lambda(a)=a_0\cup \bigcup_{n\in N\setminus 1} \{a_0+q_n\}$$
Intuition 4
For a continued fraction we have:
$$ 
a_0+\underset{n\in N\setminus 1 }{\LARGE\mathbb{K}}\frac{1}{a_n} = a_0 +\sum_{n\in N\setminus 1}\frac{(-1)^n}{q_n\cdot q_{n-1}}
$$
and the set $\lambda(a)$ captures all details of the sequence $a$.
The challenge is to explicitly define the arithmetic of continued fraction (analogous to Definition 13, 14, 15) and then re-use Definitions 3, 4, 5 to complete the construction.
http://mathworld.wolfram.com/RegularContinuedFraction.html
Update 2
I've added the extension of the overline operation for completeness sake. A more detailed version of this note is available here.
A: Let $\Bbb Z^-= (\Bbb N\setminus \{0\})\times \{M\}$ where $M$ is some (any) set such that $\Bbb Z^-$ is disjoint from $\Bbb N.$ And let $\Bbb Z=\Bbb N \cup \Bbb Z^-.$ Extend the operations $+$ and $\times$ from $\Bbb N$ to $\Bbb Z$ as follows:
$0+z=z+0=z$ and $0\times z=z\times 0=0$ for all $z\in \Bbb Z^-$.
If $x=(m, M)$ and $y=(n,M)$ belong to $\Bbb Z^-$ then $x+y=(m+n,M)$ and $x\times y=m\times n.$
If $x=(m,M)\in \Bbb Z^-$ and $0\ne n\in \Bbb N$ then $x\times n=n\times x=(m\times n, M).$
If $x=(m,M)\in \Bbb Z^-$ and $n\in \Bbb N$ then 
$\quad (i)\;\; x+n=n+x= n-m$ if $n\ge m$
$\quad (ii)\; x+n=n+x=(m-n,M)$ if $n<m.$
A: Yes (and the same argument goes for the other number sets).
Any construction of $\mathbb{Z}$ comes equipped with an embedding $i : \mathbb{N} \hookrightarrow \mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$\mathbb{Z}' = \mathbb{N} \cup (\mathbb{Z} \setminus i[\mathbb{N}])$$
where $i[\mathbb{N}] = \{ i(n) \mid n \in \mathbb{N} \}$ is the image of $i : \mathbb{N} \hookrightarrow \mathbb{Z}$.
Evidently $\mathbb{N} \subseteq \mathbb{Z}'$ and the arithmetic operations of $\mathbb{N}$ are preserved in $\mathbb{Z}'$. And indeed, $\mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $\mathbb{Z}' \to \mathbb{Z}$ given by
$$n \mapsto \begin{cases} i(n) & \text{if } n \in \mathbb{N} \\ n & \text{if } n \in \mathbb{Z} \setminus i[\mathbb{N}] \end{cases}$$
[Slight caveat: if your construction of $\mathbb{Z}$ already contains some natural numbers for whatever reason, replace $\mathbb{Z} \setminus i[\mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(\mathbb{Z} \setminus i[\mathbb{N}]) \times \{ 0 \}$.]
A: In this section we construct a commutative group. The second section develops the theory that uses this group in its presentation of the integers.
In set theory our starting point is the set of natural numbers $\Bbb N = \{0,1,2,\dots\}$ under addition $\text{+}$. Recall that any natural number $m \gt 0$ has a unique predecessor denoted by $m - 1$.
Define the set
$$\quad \Bbb Z = \{ (x,y) \in \Bbb N  \times \Bbb N \; | \; y = 0 \text{ or } x = 0\}$$
Define the bijective transformation $\sigma: \Bbb Z \to \Bbb Z$ by
$$
\sigma(x,y) = \left\{\begin{array}{lr}
        (x+1,y)\, \;\;\;\text{ |} & \text{for } y = 0\\
        (x,y-1) \,\;\;\; \text{ |} & \text{for } y \gt 0
        \end{array}\right\}
$$
Recall that if $n \in \Bbb N$ the composition of $\sigma$ with itself $n$ times is denoted by $\sigma^n$, where $\sigma^0 = id_{\Bbb Z}$.
Let $\tau$ be the inverse mapping of $\sigma$. 
If $\beta$ is in the group $\mathcal Z$ of bijective transformations generated by $\sigma$ then
$$ [\exists!\, m \in \Bbb N \text{ such that } \beta = \sigma^m] \;\text{ XOR } \;[\exists!\, m \in \Bbb N \text{ such that } m \gt 0 \land \beta = \tau^m ]$$
In the next section we show how the commutative group $(\mathcal Z,\circ)$ can be regarded as
the set of all integers.
If the reader wants to skip the next section, they should note that the mapping 
$$ m \mapsto \sigma^m$$
is an injective morphism of $(\Bbb N,+)$ into $(\mathcal Z,\circ)$.

To supply the arguments constructing the integers, you'll need the following prerequisites:

A basic knowledge of elementary algebra and the following result,
Theorem 1: Let $X$ be a set with $x_0 \in X$ and $\psi: X \to X$ be any function. Then there exist one and only one function $\rho: \Bbb N \to X$ satisfyings
$\tag 1 \rho(0) = x_0$
$\tag 2 \forall n \in \Bbb N, \; \rho(n+1) = \psi(\rho(n))$
See, for example, the wikipedia article recursive definition.

Using theorem 1 and induction, it can be demonstrated that if $f: X \to X$ is any function then
$\tag 3 f \mapsto f^n, \; n \in \Bbb N$
is well-defined, where $f^0 = id_X$ and  $f^{n+1} = f^{n} \circ f$ and that
$\tag 4 f^{m+n} = f^{m} \circ f^{n} \; \forall m,n \in \Bbb N$
If $f: X \to X$ is a bijection let $h$ denote the inverse of $f$. For $a \in X$, define
$\tag 5 X_a = \{x \in X \, \colon \, \exists k \in \Bbb N \text{ such that } [\,x = f^k(a)\,] \lor [\,x = h^k(a)\,]\}$
A simple argument show that $f(X_a) = X_a$ and $h(X_a) = X_a$
(when $X_a$ is finite, $f_{|X_a}$ correspond to a cyclic permutation).
Theorem 2: There exist a set $\Bbb Z$ with $\Bbb N \subset \Bbb Z$ and a bijective transformations $\sigma$ on $\Bbb Z$ such  that
$\tag 6 \sigma(n) = n + 1 \text{ for } n \in \Bbb N$
$\tag 7 \text{If } [\,S \subset \Bbb Z\,] \land [\,S \ne \emptyset \,]\land [\,\sigma(S) = S \,]\text{ Then } S = \Bbb Z$
If $\tau$ is the inverse of $\sigma$, the group $\mathcal Z$ generated by $\sigma$ and $\tau$ is commutative.
Also, given any $m \in \Bbb Z$ there exist one and only one $\gamma \in \mathcal Z$ such that $\gamma(0) = m$.
In this way, the group structure on $(\mathcal Z,\circ)$ can be bijectively transported to the set $\Bbb Z$ creating the additive group $(\Bbb Z,+)$ and 
having the property that $(\mathbb{N},+)  \hookrightarrow (\mathbb{Z},+)$ is a morphism.
Finally, the set $\Bbb Z$ is unique up to a bijective correspondence, in alignment with a (natural) commutative diagram.
We leave it to the interested reader to 'lift multiplication' from $\Bbb N$ to $\Bbb Z$.
