Let $R$ be a Dedekind domain and $K$ its field of fractions. Suppose $S$ is a ring with $R \subseteq S \subseteq K$. Let $P$ be a nonzero prime ideal of $S$ and set $p=P \cap R$. It is simple to show that $p$ is a nonzero prime of $R$. I want to show that $S_P=R_p$.
$R_p \subset S_P$: If $r/s \in R_p$, where $r \in R$ and $s \notin p=P \cap R$. If $s \in P$, then as $s \in R$, we have $s \in P \cap R=p$, a contradiction. But if $s \notin P$, then $r/s \in S_P$.
However, the other direction eludes me. If $s/q \in S_P$, $s \in S, q \notin P$, then certainly $q \notin p$. However, we could have $s \in S \setminus R$ and I am unsure of how to proceed.
Start by looking the example $R = \mathbb{Z}$. Take $S = \mathbb{Z}[1/2]$ and $P = (3)$. In the case where you're having trouble with the proof, you might have, say, $s = 1/2$ and $q = 5/4$. Clearly the solution in this case is to clear denominators: $s/q = 4s/4q = 2/5$.
So, in general, I would look for an element in $t \in R \backslash P$ such that both $ts, tq \in R$.