Let $G$ be a free group on a finite alphabet $A$, and $R = \text{End}(G)$. Then we can concatenate endmorphisms pointwise to get another one. But endomorphisms preserve concatenation so this must mean that $f \circ (g \cdot h) = (f\circ g) \cdot (f\circ h)$ and similarly for $(g\cdot h) \circ f$. But, the multiplicative identity would be $\text{id} \in R$, and for certain when $|A| \geq 2$, $G$ is not a commutative group.
So I don't know what is going on. I would like there to be some sort of ring-like structure, but this contradicts known observations.
This is where you go wrong. You don’t necessarily get another one unless the group is commutative.
For suppose $f \cdot g$ is a group homomorphism. Then $(f(a) g(a))^{-1} = g(a)^{-1} f(a)^{-1}$ by the normal group laws, but because $f \cdot g$ is an endormorphism, we also have $(f(a) g(a))^{-1} = f(a^{-1}) g(a^{-1}) = f(a)^{-1} g(a)^{-1}$. Then we have that $f(a)$ and $g(a)$ commute for all $a$. Obviously, this is not true in general.
For an Abelian group, we do indeed get a not-necessarily-commutative unital ring.