We prove that if $M$ is faithful $R$-module and $M$ is $S$-Noetherian, then $R$ is $S$-Noetherian.
Fix $s_0\in S$ such that $Ms_0\subseteq N$ for some finitely generated $R$-submodule of $M$. Write
$$K = \{x\in R\,|\,s_0x=0\}$$
Note that $\mathrm{Ann}_R(N) \subseteq K$. Indeed, if $Nx=0$, then $0 = Ms_0x$ and hence (by the fact that $M$ is faithful) $s_0x = 0$. Thus $x\in K$. Pick generators $n_1,...,n_k$ of $N$ and consider a morphism
$$\phi:R \rightarrow \bigoplus_{i=1}^kn_iR = L$$
given by $R\ni x\mapsto (n_ix)_{1\leq i\leq k}\in \bigoplus_{i=1}^kn_iR$. Clearly $L$ is $S$-Noetherian as a finite direct sum of $S$-Noetherian modules $n_iR\subseteq M$ for $1\leq i\leq k$. Now the kernel of $\phi$ is $\mathrm{Ann}_R(N)$. This implies that $R/\mathrm{Ann}_R(N) = R/\mathrm{ker}(\phi)$ is $S$-Noetherian as an $R$-submodule of $S$-Noetherian module $L$.
Now pick any ideal $I\subseteq R$. By bold statement above there exist $s\in S$ and a finitely generated ideal $J$ of $R$ such that
$$Is\subseteq J+\mathrm{Ann}_R(N)\subseteq J+ K$$
Then
$$Iss_0 \subseteq Js_0$$
Now $ss_0\in S$ and $Js_0$ is a finitely generated ideal of $R$. Thus $R$ is $S$-Noetherian.
Remark.
$K = \mathrm{Ann}_R(s_0)$
where $s_0$ is treated as an element of $R$-module $R$.