Let $R$ be a (commutative) integral domain. Denote the field of fractions of $R$ by $Q(R)$.
Assume that $S \subseteq R$ is a subring of $R$, denote the field of fractions of $S$ by $Q(S)$ and further assume that: $Q(S) \cap R=S$ (of course, we always have $Q(S) \cap R \supseteq S$).
I wonder if the following claim is true:
If $\sum_i \tilde{s_i} r^i \in R$, where $\tilde{s_i} \in Q(S), r \in R$, then necessarily $\tilde{s_i} \in R$ or $r \in S$. In other words, if $\sum_i \tilde{s_i} r^i \in R$, where $\tilde{s_i} \in Q(S), r \in R$, then necessarily $\tilde{s_i} \in S$ or $r \in S$.
Remark: Probably, the answer depends on the specific $R$ and $S$ or on connections between them, for example flatness of $S \subseteq R$ etc.
Edit: An easy counterexample is presented in the answer to this question: $S=k[y^2], R=k[y], \tilde{s_1}=\frac{1}{y^2}, r=y^3$; indeed, $\tilde{s_1}r=\frac{1}{y^2}y^3=y \in k[y]$ with $\tilde{s_1}=\frac{1}{y^2} \notin k[y]$ and $r=y^3 \notin k[y^2]$.
Therefore, I would like to ask:
Are there additional conditions guaranteeing the following property to be satisfied: If $\sum_i \tilde{s_i} r^i \in R$, where $\tilde{s_i} \in Q(S), r \in R$, then necessarily $\tilde{s_i} \in S$ or $r \in S$.
Examples for additional conditions:
(a) $S \subseteq R$ is separable. (b) The only invertible elements of $S$ and $R$ are $k^*=k-\{0\}$, where $k$ is the base field.
(In the counterexample, $S \subseteq R$ is integral, so we see that integrality will not help. $R=k[y]=k[y^2][y]=S[y]$, so being simple will not help also).
Thank you very much!