Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 \in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a \in A$.
From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.
So you only have problems with inverses? Then here you go:
first of all: $\mathrm{Mod}_A(M,N)\subset \mathrm{Set}(M,N)$, where $\mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $\mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $\mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $\mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M \to N$ is clearly an $A$-homomorphism, hence $\mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M \to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: M\to N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M \to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$ which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 \in A$.