A few more details regarding the last sentence. If $G$ is a countable group and $\tau \in \mathcal{T}^0_G$ is a complete extension of $A_G$, then $\tau$ corresponds to a ring with underlying additive group $G$ if and only if there is no single formula $\phi(v_1)$ such that $\tau \vdash \phi(v_1) \to v_1 \neq a$ for every $a \in G$. Of course, this happens exactly when $(\forall v_1 \lnot\phi(v_1)) \in \tau$ or $\phi(a) \in \tau$ for some $a \in G$. So the points of $\mathcal{T}^0_G$ that correspond to types that contain one of these for each formula $\phi(v_1)$. Since $G$ is countable, this is always a $G_\delta$ subset of $\mathcal{T}^0_G$ and therefore it is always a nice Polish subspace of $\mathcal{T}^0_G$.
Since you're interested in the action of $\mathrm{Aut}(G)$, let me add that $\mathrm{Aut}(G)$ acts on $\mathcal{T}^0_G$ in a natural way and that the $G_\delta$ subset described above is invariant under this action. However, since $\mathcal{T}^0_G$ is compact, there may be much to gain in considering the action on the whole space.
Here is the setup for your case for $0$-types. Let $\mathcal{L}_G$ be the language of rings augmented with a constant for each element of the additive group $G$. Let $A_G$ be the set of purely additive (have no mention of multiplication or the multiplicative identity) sentences of $\mathcal{L}_G$ that are true in $G$. This is a partial $0$-type and the space $\mathcal{T}_G^0$ of all (complete) $0$-types extending $A_G$ is a lot like what you describe. Indeed, if $a,b,c \in G$ then $a\times b = c$ is a sentence of $\mathcal{L}_G$ which determines a basic clopen set $\lbrace p \in \mathcal{T}_G^0 : a \times b = c \in p\rbrace$. This is a little finer than the space $\mathcal{R}_G$ you describe since elements of $\mathcal{R}_G$ only encode the truth for quantifier-free sentences of $\mathcal{L}_G$. The advantage of using spaces of types is that they have been extensively studied in model theory. Beyond $0$-types, you can consider the spaces of $k$-types, which instead of sentences use the broader class of formulas of $\mathcal{L}_G$ with free variables among $v_1,\dots,v_k$. Together, these spaces of types give a very nice understanding of the situation you're looking at.
Note that there are some types that are not necessarily compatible with the underlying additive group being precisely $G$. For example, the $1$-types that extend $\lbrace v_1 \neq a : a \in G \rbrace$ are incompatible with such rings. Nevertheless, if $R$ is a ring with underlying additive group $G$, then $R$ corresponds to the $0$-type $\lbrace \phi : R \vDash \phi\rbrace$, every element $a \in R$ corresponds to the $1$-type $\lbrace \phi(v_1) : R \vDash \phi(a)\rbrace$, and so on. The Omitting Types Theorem lets you know which types can correspond to rings whose underlying groups are precisely $G$.