Your isomorphism holds in general. Namely if $N$ is a $R$-module with the action of a finite group $G$ and if $f\in R$, then $R_f\otimes N^G=(N_f)^G$ as submodules of $N_f$. Indeed $R_f\otimes N^G\subset (N_f)^G$ is clear. Conversely, suppose $v=x/f^n\in (N_f)^G$. For any $\sigma\in G$, we have $\sigma(x)/f^n=x/f^n$. So $f^m(\sigma(x)-x)=0$ for some $m\in \mathbb N$ (depending on $\sigma$). As $G$ is finite, there is an $m_0$ big enough such that $f^{m_0}(\sigma(x)-x)=0$ for all $\sigma\in G$. Then $f^{m_0}x\in N^G$ and $v=f^{m_0}x/f^{n+m_0}\in R_f\otimes N^G.$
Of course, if $R$ is a domain $N$ has no $R$-torsion ($fx=0$ with $f\in R$ and $x\in N$ implies $f=0$ or $x=0$), then the proof is much simpler and we don't need the finiteness of $G$.
Generalization Let $N, G$ be as above. Let $R\to R'$ be a flat extension. Then $G$ acts naturally on $R'\otimes N$ (trivially on $R'$) and we have
$ R'\otimes_R N^G = (R'\otimes_R N)^G$ as submodules of $R'\otimes N$.
Proof: For any $\sigma\in G$, denote by $\sigma_N : N\to N$ the image of $\sigma$ in $\mathrm{Aut}_R(N)$. We have $N^G=\cap_{\sigma\in G} \ker(\sigma_N - 1).$ As $R'$ is flat and $G$ is finite, $ R'\otimes \cap_{\sigma\in G} \ker(\sigma_N - 1)=\cap_{\sigma\in G} (R'\otimes \ker(\sigma_N-1))=\cap_{\sigma\in G} \ker(\sigma_{R'\otimes N} - 1)=(R'\otimes N)^G.$ Hence $R'\otimes N^G=(R'\otimes N)^G$.