There are several canonical group structures that can be equipped naturally on a group ; I am going to list the ones that I know that are interesting, perhaps other people can add some to the list.

Finite abelian groups are isomorphic to product of finite cyclic groups so you can naturally form a ring from a direct product of finite rings. The same goes for finitely generated abelian groups ; you can deduce a ring structure from the products of the underlying natural rings.

One ring structure that is always there will be the trivial ring structure, i.e. just add a multiplication that means nothing ; it sends everything to the identity (of $G$). Clearly it respects all the ring axioms and it gives you a ring (without a *ring* identity, though). It is not very interesting in such, because it means "we do have a ring structure on $G$, but it is not telling us anything at all about $G$".

Another ring structure which does not always have its underlying abelian group isomorphic to $G$ but is canonically defined by $G$ is the *endomorphism ring* of $G$, i.e. let
$$
R = \{ \phi : G \to G \, | \, \phi \text{ is a group endormorphism of } G \text{ to } G \}.
$$
Equip a ring structure on $R$ by saying that $(R,+, \circ)$ is a ring by defining the following : the endomorphism $\phi_1 + \phi_2$ is defined as
$$
(\phi_1 + \phi_2)(g) = \phi_1(g) + \phi_2(g)
$$
and
$$
(\phi_1 \circ \phi_2)(g) = \phi_1(\phi_2(g)).
$$
If $G$ is abelian, then you can readily see that the operation $+$ gives you an abelian group on $R$, because the inverse of the endomorphism $\phi$ will just be the endomorphism that sends $g$ to $-\phi(g)$ instead of $\phi(g)$, so that $\phi + (-\phi) = 0_R$. Commutativity follows from the fact that $G$ is abelian and the other properties follow naturally. The fact that the $\circ$ operation gives you a ring structure comes from the fact that endomorphisms are in particular homomorphisms, so that
$$
(\phi \circ (\psi_1 + \psi_2))(g) = \phi( \psi_1(g) + \psi_2(g)) = (\phi \circ \psi_1)(g) + (\phi \circ \psi_2)(g)
$$
In particular, the identity of the ring $R$ would be the identity endomorphism of $G$.

I know it is not exactly what you asked for (in this last part), but I didn't know much about it at first when I saw it and I believed it was interesting, so I just thought you might wanna take a look.

Hope that helps,