Ring structure with underlying abelian group $G$ If we start with an abelian group $G$, a ring structure with underlying group $G$ is in particular a map $\mu: G \otimes G \to G$, which by adjunction defines a map $\tilde{\mu}:G \to \operatorname{Hom}_{\mathbb{Z}}(G,G)$. 
But every map $\mu:G \otimes G \to G$ will not give an associative unital multiplication structure. In order to get associativity and unity in terms of the map $\tilde{\mu}$ (i.e. in order to piggyback on the ring structure of $\operatorname{End}_{\mathbb{Z}}(G)$), it seems we should require $\tilde{\mu}$ to be injective, and that $\operatorname{Id}_G$ be in $\operatorname{Image}(\tilde{\mu})$. Am I correct that those are the right hypotheses on $\tilde{\mu}$, and/or is there a cleaner way to do this in terms of $\tilde{\mu}$ only?
These questions (1) (2) raise related issues.
 A: Given a map $\varphi : G \to \mathrm{End}_{\mathbb Z}(G)$ (denote $\varphi(g)$ by $\varphi_g$) satisfying 
$$
\forall s,t \in G, \quad \varphi_s \circ \varphi_t = \varphi_{\varphi_s(t)}, \quad \exists 1_G \in G \quad \text{ with } \quad \varphi_{1_G} = \mathrm{id}_G,
$$
you get a ring structure $(G,+,\cdot)$ on your abelian group $(G,+)$ ; multiplication is defined by $s \cdot t \overset{def}= \varphi_s(t)$. Bilinearity follows from the tensor-hom adjunction and associativity follows from the first equation since 
$$
s \cdot (t \cdot u) = \varphi_s(\varphi_t(u)) = \varphi_{\varphi_s(t)}(u) = (s \cdot t) \cdot u. 
$$ 
Up to now we have a (non-unital) ring ; the fact that $1_G \cdot s = s$ for all $s \in G$ is trivial, which shows that such a ring has a left-identity element ; it is not clear that it will have a right-identity element. The extra condition that you want to have a right identity element is that $\varphi_s(1_G) = s$ for all $s \in G$, and it doesn't follow automatically.
Added : Thanks to Omar Antolín-Camarena in the comments, we have the example of the non-commutative ring 
$$
\left\{ \left. \begin{pmatrix} x & y \\ x & y \end{pmatrix} \, \right| \, x,y \in \mathbb R \right\}
$$
It is not commutative, as is seen by the multiplication rule 
$$
\begin{pmatrix} x & y \\ x & y \end{pmatrix} \begin{pmatrix} z & t \\ z & t \end{pmatrix} = \begin{pmatrix} (x+y)z & (x+y)t \\ (x+y)z & (x+y)t \end{pmatrix}
$$
and a left-identity is given by $\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$. Setting $z=1$ and $t=0$, the product above gives $\begin{pmatrix} x+y & 0 \\ x+y & 0 \end{pmatrix}$, so that our left-identity is not a right-identity. In particular, the corresponding map $\varphi$ is not injective ; there is more than one left-identity, namely $\begin{pmatrix} \lambda & 1-\lambda \\ \lambda & 1-\lambda \end{pmatrix}$ for $\lambda \in \mathbb R$.
Hope that helps,
