@Arrow not sure that answers your question, but one way to think of the group (or more generally monoid) algebra is this: suppose $R$ is commutative (you want that to talk about $R$-algebras anyway, I think), then the free functor $\mathbf{Set} \to R\textrm{-}\mathbf{Mod}$ is strong monoidal where we consider the monoidal structure on $\mathbf{Set}$ given by products and on $R\textrm{-}\mathbf{Mod}$ by tensor products,
since there is a natural bijection $R^{(X)} \otimes_R R^{(Y)} \cong R^{(X \times Y)}$ satisfying the necessary coherence axioms.

@Arrow once you have the adjunction between monoids and $R$-algebras, you can use that to construct the arrow $RG \to \operatorname{Hom}_{R\textrm{-}\mathbf{Mod}}(R^{(X)},R^{(X)})$ from an arrow of monoids $G \to \operatorname{Hom}_{\mathbf{Set}}(X,X)$:
Just take the compose $G \to \operatorname{Hom}_{\mathbf{Set}}(X,X)$ with the monoid homomorphism induced on endomorphism monoids by the free module functor $\operatorname{Hom}_{\mathbf{Set}}(X,X) \to \operatorname{Hom}_{R\textrm{-}\mathbf{Set}}(R^{(X)},R^{(X)})$ and then apply the adjunction to get an $R$-algebra morphism $RG \to \operatorn…