Prove that if $A$ is a ring and $G$ is a group, then the map $$\text{Hom}(\mathbb Z[G],A) \to \text{Hom}(G,\mathcal U(A))$$ which sends $f \rightarrow f|_G$ is a bijection.
First of all, I am having some problem with notation here (or maybe I don't understand the definition of group ring): Shouldn't it say $1.G$ instead of $G$?
I could show that the map "makes sense",i.e, that $f|_{1.G}$ is a morphism from $G$ to $\mathcal U(A)$. I couldn't show that $\psi$ is injective and surjective.
For injectivity, I need to show tht if $\psi(f)=\psi(g) \implies f=g$, this is equivalent to show that if $f|_G=g|_G$, then $f=g$.
For surjectivity, given $h \in \text{Hom}(G,\mathcal U(A))$, I need to find a morphism $f \in \text{Hom}(\mathbb Z[G],A)$ such tht $f|_G=h$
Any suggestions would be appreciated.
The notation $f|_G$ stands for what you called $f|_{1.G}$. The idea is that $G$ is embedded inside $\mathbb{Z}[G]$ via the map $g\in G \mapsto 1.g \in \mathbb{Z}[G]$. So the notation $f|_G$ means restricting to $G$ identified with this subgroup of $\mathbb{Z}[G]$. I think this notation is standard.
Also it's common to write $A^\times$ for the units of the ring $A$.
Anyway the point is that if you have a map $h: G \rightarrow A^\times$, then you can simply make a map $\eta: \mathbb{Z}[G] \rightarrow A$ by setting $\eta(n.g) = n (h(g))$, where on the right hand side you have $n$ times the ring element $h(g)\in A$. Similarly you can define $h$ on all other elements of $\mathbb{Z}[G]$ by the requirement that $h(x+y)=h(x)+h(y)$. It's not hard to show $\eta|_G = h$.
For injectivity, the point is that a ring homomorphism $\eta: \mathbb{Z}[G]\rightarrow A$ is completely determined by its values on $G \subset \mathbb{Z}[G]$, because these elements generate the ring $\mathbb{Z}[G]$. In other words if you know $\eta(g_1)$ and $\eta(g_2)$, then you can find $\eta(9g_1 + 6g_2)$. This shows $\eta|_G$ has all the information necessary to recover $\eta$.