$\newcommand{\Z}{\mathbb{Z}}\newcommand{\F}{\mathbb{F}}\newcommand{\N}{\mathbb{N}}$Here's a remark on Martin's answer, which is unfortunately too long to post as a comment.
The point I want to mention is that the simplification noted in Martin's answer, where one considers tensor products not in $\mathsf{CMon}$, but rather in the category of commutative monoids with zero, comes from considering tensor products not over $\N$, but over the field with one element $\F_1$.
Firstly, note that we may build $\mathsf{CMon}$ by first considering $\N$-modules, which are just monoids, and then considering commutative monoids in that category, giving us commutative monoids again, i.e. $\mathsf{CMon}\cong\mathsf{CMon}(\mathsf{Mod}_{\N})$.
We can, however, start by considering not $\N$-modules, but $\F_1$-modules. These are usually thought to be pointed sets (i.e. we define $\mathsf{Mod}_{\F_1}\cong\mathsf{Sets}_*$). Now, the category $\mathsf{Sets}_*$ admits a monoidal structure $\otimes_{\F_1}$ obtained by defining
$$
(X,x_0)\otimes_{\F_1}(Y,y_0)=(X\times Y/{\sim},[(x_0,y_0)]),
$$
where ${\sim}$ is the equivalence relation on $X\times Y$ obtained by declaring
\begin{align*}
(x,y_0) &\sim (x_0,y_0),\\
(x_0,y) &\sim (x_0,y_0),
\end{align*}
for each $x\in X$ and each $y\in Y$. The monoidal unit of this tensor product is then $\mathbb{F}_1=\{0,1\}$.
Now, a commutative monoid in $(\mathsf{Mod}_{\F_1},\otimes_{\F_1},\F_1)$ is precisely a commutative monoid with zero! Indeed, it will be a triple $((A,0_A),\mu,\eta)$ consisting of
- A pointed set $(A,0_A)$;
- A morphism of pointed sets
$$\mu\colon(A,0_A)\otimes_{\F_1}(A,0_A)\to(A,0_A),$$
being equivalently given by a map of sets $\mu\colon A\times A\to A$, the multiplication of $A$, which is $\F_1$-bilinear in that we have
\begin{align*}
0_Aa &= 0_A,\\
a0_A &= 0_A;
\end{align*}
- A morphism of pointed sets
$$\eta\colon(\F_1,0)\to(A,0_A),$$
picking an element $1_A$ of $A$, its unit.
So, in a sense, commutative monoids with zero are $\F_1$-algebras, and considering the tensor product of the $\F_1$-algebra $\Z$ with itself as an $\N$-algebra will give the wrong result, the complicated tensor product in Martin's answer.
Finally, here's a picture I drew of the tensor product $\N\otimes_{\F_{1}}\N\cong(\N\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$, where $\mathbb{N}$ is the multiplicative monoid of natural numbers:
Here the blue circles depict the element $(2$, $1$, $4$, $1$, $2$, $3$, $4$, $2$, $2$, $1$, $2$, $3$, $2$, $1$, $4$, $4$, $1$, $2$, $2$, $2$, $2$, $4$, $3$, $1$, $3$, $2$, $1$, $3$, $4$, $2$, $2$, $2$, $3$, $3$, $4$, $2$, $1$, $2$, $1$, $2$, $2$, $2$, $\ldots)$ of $\N\otimes_{\F_{1}}\N\cong(\N\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$, represented as a counterclockwise spiral.
(Note that the factors of $\mathbb{N}$ in the above direct sum carry the multiplicative monoid structure, not the additive one!)
We can then picture $\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$ in a similar way, adding negative numbers to the above image.
