Recovering $\mathsf{Ab}$ from $\mathbb{Z}$-mod in a closed symmetric monoidal category Given the usual definition of a module over a ring it is trivial to show that a $\mathbb{Z}$-module is an abelian group (it is just by definition).
But my question concerns recovering this idea in a more abstract setting.
Let $$(\mathcal{C},\otimes,1)=(\mathsf{Set},\times,\{*\})$$ be the closed symmetric monoidal category of sets with the cartesian product.
Then we have the notion of a commutative monoid in $\mathcal{C}$: an object $A\in\mathcal{C}$ along with morphisms $\mu\colon A\times A\to A$ and $\eta\colon 1\to A$ satisfying certain commutative diagrams.
If we have such a monoid $A$ then we can define a module over $A$: an object $M\in\mathcal{C}$ along with an action $\sigma\colon A\times M\to M$ satisfying certain commutative diagrams.
Now, by closedness of $\mathcal{C}$, we know that specifying an action $\sigma\colon A\times M\to M$ is equivalent to specifying a morphism $\varphi\colon A\to\mathrm{Hom}_{\mathsf{Set}}(M,M)$ given by currying.
That is, $$\varphi(a)(-)=\sigma(a,-).$$
It is just a matter of diagram chasing to then show that the axioms that make $\sigma$ an action give us certain constraints on $\varphi$.
In particular, they ensure that $\varphi$ is a morphism of monoids (where the monoid structure on $\mathrm{Hom}_{\mathsf{Set}}(M,M)$ is given by composition, and $\eta\colon 1\to\mathrm{id}_M$).
Taking $A=\mathbb{Z}$ (since $\mathsf{Comm}(\mathsf{Set})=\mathsf{CMon}$, the category of commutative monoids in the usual sense), we see that $M\in\mathsf{Set}$ can always be seen as a $\mathbb{Z}$-module by taking the trivial action $\sigma\colon(n,x)\mapsto x$.
This corresponds to having $\varphi\colon n\mapsto\mathrm{id}_M$.
More generally though, an action $\sigma$ determines a monoid morphism $\varphi\colon\mathbb{Z}\to\mathrm{Hom}_\mathsf{Set}(M,M)$ satisfying


*

*$\varphi(m+n)=\varphi(m)\circ\varphi(n)$;

*$\varphi(0)=\mathrm{id}_M$.


My (simple) question: how does this allow us to consider some $\mathbb{Z}$-module $M$ as an abelian group - what is our binary operation $M\times M\to M$?
Edit: thinking about it more, it seems like it might be more logical to think of $\mathbb{Z}\in\mathsf{CMon}$ as a multiplicative monoid, since the natural $\mathbb{Z}$ action on a group comes from $$nx:=\underbrace{x+\ldots+x}_{n\text{ times}}$$
in which case we would have $\varphi(mn)=\varphi(m)\circ\varphi(n)$ and $\varphi(1)=\mathrm{id}_M$, but I still can't quite piece it all together.
Edit (follow-up question): if we set $(\mathcal{C},\otimes,1)=(\mathsf{Ab},\otimes,e)$ the be the category of abelian groups with the usual tensor product then, by definition, we see that, since $\mathbb{Z}$ is a ring, and $\mathsf{Comm}(\mathcal{C})$ is the category of commutative rings, $\mathbb{Z}$-mod is exactly the category of abelian groups again.
So do we get different $\mathbb{Z}$-mods depending on how much structure of $\mathbb{Z}$ we use?
 A: Your first construction is a category of sets a set-monoid acts on.
Its objects are called monoid actions, and are exactly what you wrote down - either sets equipped with a nice operation $A × M → M$, or a way to interpret elements of $A$ as transformations of $M$ (ie. a nice function $A → \mathrm{End}(M))$. These things certainly don't come automatically equipped with a natural binary operation for any monoid $A$.
What is usually meant by $ℤ$-modules are on the other hand the category of abelian groups the $(\mathrm{Ab}, ⊗)$-module $ℤ$ acts on, ie. exactly your second construction.
To make your last comment on the amount of structure precise, note that the forgetful functor $U : \mathrm{Ab} → \mathrm{Set}$ is monoidal, and every monoidal functor sends monoids to monoids, and furthermore lifts to a functor between the categories of modules over those monoids.
In particular, $(\mathrm{Ab}, ⊗)$-module/ring $ℤ$ gets sent to $(\mathrm{Set}, ×)$-monoid $(ℤ, \cdot)$, and every abelian group gets sent to its underlaying $ℤ$-action.
