Modules as morphisms to endomorphism rings An $A$-module $M$ may be thought of as a (surjective) ring homomorphism $f: A \to E(M)$, where $E(M)$ is a ring of group endomorphisms of $M$. Then $am = f(a)(m)$.
Is there any more to this correspondence? Is there a way to frame module homomorphisms, bilinear products,... in this view?
 A: This is a very general question and can not be answered in a few words. So, may be let me describe one aspect. If $R$ is a commutative ring and $M$ an $R$-module, giving an $R$-algebra homomorphism $A\to \operatorname{End}_R M$, where $A$ is an $R$-algebra makes $M$ into an $A$-module. In the example you write above, $R=\mathbb{Z}$.
Now, let me look at a special case, where $R$ is a field and $A$ is finite dimensional commutative $R$-algebra. Then, we have natural perfect pairing $A\times \operatorname{Hom}_R(A,R)\to R$. But, easy to see that $\operatorname{Hom}_R(A,R)=\omega_A$ has a natural $A$-module structure using the above. Then, a whole lot of duality can be worked out using $\omega_A$ for $A$-modules, imitating what happens for vector spaces over $R$. It is very powerful, as duality is for vector spaces. 
A: Yes, of course, but it's not that useful. Suppose you have two ring homomorphisms $f\colon A\to E(M)$ and $g\colon A\to E(N)$. Suppose you also have a group homomorphism $h\colon M\to N$ is a module.
For each $a\in A$, you have $f(a)\colon M\to M$ and $g(a)\colon N\to N$; so you can form the square
$$\require{AMScd}
\begin{CD}
M @>f(a)>> M \\
@VhVV @VVhV \\
N @>g(a)>> N
\end{CD}
$$
and $h$ is a module homomorphism precisely if, for every $a\in A$, the diagram commutes.
Note that there's generally no way to define a ring homomorphism $E(M)\to E(N)$ starting from a group homomorphism $h\colon M\to N$. For instance, if $M=\mathbb{Q}$ and $N=\mathbb{Q}/\mathbb{Z}$, there is no ring homomorphism $E(\mathbb{Q})\to E(\mathbb{Q}/\mathbb{Z})$, because, looking at the additive structure, $E(\mathbb{Q})\cong\mathbb{Q}$ is a divisible group, whereas $E(\mathbb{Q}/\mathbb{Z})$ is a reduced group (it has no divisible subgroups)1. So the only abelian group homomorphism $E(\mathbb{Q})\to E(\mathbb{Q}/\mathbb{Z})$ is zero.
On the other hand both groups are $\mathbb{Z}$-modules.
You can't even go the other way around: consider the canonical injection $\mathbb{Z}\to\mathbb{Q}$: there is no ring homomorphism $E(\mathbb{Q})\to E(\mathbb{Z})$.

1 The ring $E(\mathbb{Q}/\mathbb{Z})$ is the product for $p$ running through all primes of the rings of $p$-adic integers, which are known to be reduced groups.
