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,