Let $\mathcal{O}$ be a Dedekind domain with ring of fractions $K$. Given a cofinite set of primes $X\subseteq \operatorname{Spec}(\mathcal{O})$, we can form the localization with respect to the multiplicatively-closed set $S = \mathcal{O}\setminus \bigcup_{\mathfrak{p}\in X}\mathfrak{p}$:
$$\mathcal{O}(X) := S^{-1}\mathcal{O} = \{ f/g\in K : f,g\in\mathcal{O}\textrm{ and }\forall\mathfrak{p}\in X, v_\mathfrak{p}(g) = 0\}.$$
We can also consider the "valuation-theoretic localization" defined as
$$\overline{\mathcal{O}}(X) := \{ x\in K : \forall\mathfrak{p}\in X, v_\mathfrak{p}(x) \geqslant 0\}.$$
Clearly $\mathcal{O}(X)\subseteq \overline{\mathcal{O}}(X)$, and I'm interested in when the reverse inclusion also holds. I can prove the reverse inclusion when $\mathcal{O}$ has torsion ideal class group, but
I wonder if there's a simpler proof and/or if the claim generalizes to arbitrary Dedekind domains.
Here's my proposed proof for the torsion case. Let $x\in\overline{\mathcal{O}}(X)$ and let
$$Y = \{\mathfrak{p}\in\operatorname{Spec}(\mathcal{O}):v_\mathfrak{p}(x) < 0\}\subseteq \operatorname{Spec}(\mathcal{O})\setminus X.$$
For each $\mathfrak{p}\in Y$, let $e_\mathfrak{p}\geqslant 1$ be the order of $\mathfrak{p}$ in the class group of $\mathcal{O}$, and let $\mathfrak{p}^{e_{\mathfrak{p}}} = (z_\mathfrak{p})$ so that
$$v_\mathfrak{q}(z_\mathfrak{p}) = \begin{cases}e_\mathfrak{p}&\mathfrak{q} =\mathfrak{p}\\0&\textrm{otherwise}\end{cases}$$
Let $e$ be the least common multiple of the $e_\mathfrak{p}$, and let
$$g := \frac{1}{\prod_{\mathfrak{p}\in Y} z_\mathfrak{p}^{v_{\mathfrak{p}}(x)e/e_{\mathfrak{p}}}}.$$
Then
$$v_\mathfrak{q}\left(g\right) = \begin{cases}-ev_\mathfrak{q}(x)&\mathfrak{q} \in Y\\0&\textrm{otherwise}\end{cases}$$
Since $v_\mathfrak{q}(x)\geqslant 0$ for $\mathfrak{q}\notin Y$, this implies $f:=x^eg\in\mathcal{O}$; and since $v_\mathfrak{q}(x)<0$ for $\mathfrak{q}\in Y$ this implies $g\in\mathcal{O}$. Thus
$$x^e = f/g\in\mathcal{O}(X).$$
But $\mathcal{O}(X)$ is a localization of the integrally-closed domain $\mathcal{O}$ and hence is itself integrally closed. Since $x$ is a root of the monic polynomial $t^e-f/g\in\mathcal{O}(X)[t]$, this implies $x\in\mathcal{O}(X)$.