If $R$ is a ring, is it abuse of notation to write $2ab$ for general elements $a,b\in R$ for $ab+ab$? I have a question regarding this answer. I am trying to prove that any Boolean ring is commutative. The answer at that link is incredibly intriguing, but there's something that bugs me slightly:
The answer starts off with $2ab$. But what is the meaning of $2ab$ exactly? I understand that rings have an underlying commutative group structure under the operation $+$, but is it safe to write the $2$ as a coefficient to denote $ab + ab$? It makes it seem as if $2$ is an element of the ring, but we never defined what the elements of $R$ are. 
There is some precedence in my textbook of using $-$ to denote the opposite of $+$, i.e., $(-a)(-b) = ab$, analogous to how we pretend there's a $-1$ out in front of the $a$ and $b$ in elementary algebra, but it honestly seems as if it's "abuse of notation" to write $2ab$ in this case where $a,b$ are elements of a ring. 
EDIT: As an addendum, my textbook does not require that rings be unital. However, the problem at hand appears in a section that says "assume $R$ is a ring with identity $1$."
 A: If a ring $R$ is unital, we adopt for natural numbers $n$ the notation
$$
n = \underbrace{1 + \cdots + 1}_{\text{$n$ times}}.
$$
If $n$ is a negative integer, we similarly define it as the sum of $-n$ copies of $-1$.
Then
$$
ab + ab = 1\cdot ab + 1\cdot ab = (1 + 1)\cdot ab = 2ab.
$$
Note that the mapping defined above gives a ring homomorphism $\mathbb Z \to R$, which tells you that with these "integers in the ring" you can do a lot of ordinary arithmetic.
Edit: If the ring is not unital, it does not make sense to talk about $2$ as an element of the ring. However, the notation $2ab = ab + ab$ still does. One way of thinking about this is as follows: even if $R$ is not unital, the endomorphism ring $\mathrm{End}(R)$ of the underlying additive group of $R$ still will be, with the identity as unit. Hence, we can interpret natural numbers not as elements of $R$, but nonetheless as a subring of the endomorphisms of the additive group of $R$. Then the above sequence of equations becomes
$$
ab + ab = \mathrm{id}_{R}(ab) + \mathrm{id}_{R}(ab) = (\mathrm{id}_{R} + \mathrm{id}_{R})(ab) = (1_{\mathrm{End}(R)} + 1_{\mathrm{End}(R)})(ab) = 2_{\mathrm{End}(R)}(ab),
$$
and we denote that last element as $2ab$.
A: No, it's not abuse.
If $M$ is a (multiplicative) monoid, you can recursively define, for $x\in M$,
$$
x^0=\mathbf{1},\qquad x^{n+1}=x^nx \tag{1}
$$
and, if $x$ is invertible, $x^n=(x^{-1})^{-n}$ when $n$ is a negative integer.
When the operation is denoted additively, the exponent is usually set besides the element:
$$
0x=\mathbf{0},\qquad (n+1)x=nx+x \tag{2}
$$
and, as before, $nx=(-n)(-x)$ if $x$ is invertible. Note that I use $\mathbf{1}$ and $\mathbf{0}$ for the neutral element, in order to avoid ambiguities.
A ring is a group with respect to addition, so the notation with multiples (2) can be freely used, hence
$$
ab+ab=2ab
$$
is fine. Here it should be interpreted as $2(ab)$. On the other hand, interpreting it as $(2a)b$ would yield the same result:
$$
(2a)b=(a+a)b=ab+ab=2(ab)
$$
and this can be generalized to any integer in place of $2$. Therefore it's not necessary to add parentheses. Also $a(2b)=2(ab)$ can be easily proved and generalized.
Note that it's not really necessary to distinguish $\mathbf{1}$ (the identity in the ring) with $1$ (the integer) and $\mathbf{0}$ (the zero in the ring) with $0$ (the integer), because
$$
0a=\mathbf{0}=\mathbf{0}a,\quad 1a=a=\mathbf{1}a
$$
It's not necessary that the ring has an identity: if it has it, denoting it by $1$ will not raise ambiguities.
If the ring $R$ has an identity (let me still denote it by $\mathbf{1}$), then there exists a unique ring homomorphism $\chi\colon\mathbb{Z}\to R$, defined by $n\mapsto n\mathbf{1}$. It's easy to prove that, for $x\in R$,
$$
nx=\chi(n)x
$$
(the left-hand side has the notation defined in (2), in the right-hand side there's the ring multiplication). So there's no ambiguity even if $2$ (or any other integer) is thought to be $2\mathbf{1}=\mathbf{1}+\mathbf{1}$. Hence usually the distinction between the two possible interpretations of $1$ is dropped. Thus it's common to see that $10=0$ in a ring of characteristic $5$ or similar identities (this is indeed a slight abuse of notation).
Note also that $-a$ can be interpreted as $(-1)a$, because both yield the same result; but it's better to think to $-a$ just as the opposite element of $a$, that is, the unique element such that $a+(-a)=\mathbf{0}=(-a)+a$.

By the way, (1) explains why, in unital rings, $\mathbf{0}^0=\mathbf{1}$ (usually written $0^0=1$).
A: You are probably more familiar with the notation $a^n$ to denote the $n$-th power of an element $a$. But how do you define it formally? Well, you do it by induction. First set $a^0 = 1$ and then, for all $n \geqslant 0$, $a^{n+1} = a^na$.
Now, just do the same thing for addition: $0a = 0$ and for all $n \geqslant 0$, $(n+1)a = na + a$. Thus $na = \underbrace{a + {}\dotsm {} + a}_{n \text{ times}}$, which is exactly what you need. 
