Suppose $R$ is a Dedekind domain with maximal spectrum $\operatorname{SpecMax}(R)$ and $X \subset \operatorname{SpecMax}(R)$ is a subset. My main question is: when can we find a multiplicatively closed subset $S \subset R$, such that $\operatorname{SpecMax}(S^{-1}R)=X$
When $X$ is finite, one can set $S= \displaystyle \bigcap_{\mathfrak{p} \in X}(R \setminus\mathfrak{p})$ and use the prime avoidance lemma together with the fact that $\operatorname{dim}(R)=1$ to show that $\operatorname{SpecMax}(S^{-1}R)=X$.
Of course, this definition of $S^{-1}R$ also makes sense if $X$ is not finite, and we still have $X \subset \operatorname{SpecMax}(S^{-1}R)$, but it is not clear to me why the right hand side might not be larger. (If $S$ doesn't work, then no multiplicatively closed subset will work, since we only need to consider saturated ones.)
We can rephrase this question in the following way: If $\mathfrak{q} \notin X$ can we find an element $q \in \mathfrak{q}$ such that $q \notin \mathfrak{p}$ for any $\mathfrak{p} \in X$. If the class group of $R$ is torsion, then we have a positive answer: suppose $\mathfrak{q}^n=(a)$, then if $a \in \mathfrak{p}$ for some $\mathfrak{p} \neq \mathfrak{q}$, we would have $\mathfrak{p} \mid \mathfrak{q}^n$ which is impossible.
My feeling is that this won't work in general if the class group of $R$ is not torsion, so the follow-up question is: What if we not only consider localizations, but arbitrary overrings (that is, rings $A$ such that $R \subset A \subset K$, where $K$ is the fraction field of $R$)?