$
\newcommand{\O}{\mathcal O}
\newcommand{\Cl}{\mathrm{Cl}}
\newcommand{\p}{\mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $\O_K$ which are representatives of the ideal classes of $\Cl(\O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $\O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $\O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $\O_{K, S}$ is a Dedekind domain, and every ideal of $\O_{K,S}$ is an extended ideal, i.e. of the form $J\O_{K,S}$ for some ideal $J \subset \O_K$.
By definition of the ideals $a_i$, there is an element $x \in K^{\times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y \O_K$ is principal (where $y \in \O_K$), since the class group is annihilated by $h$. One sees that $y \in a_i \cap \O_{K,S}^{\times}$ (because $a_i^h \subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i \O_{K,S} = (1)$
and finally
$$J \O_{K,S} = x \O_{K,S} \subset \O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $\O_{K,S}$ is equal to a localization $\Sigma^{-1}\O_K$, namely for the multiplicative set $\Sigma$ of elements $a \in \O_K \setminus \{0\}$ such that the only prime factors of $a \O_K$ belong to $S$, that is
$$\Sigma = \bigcap_{\p \not \in S} (\O_K \setminus \p) = \O_K \setminus \bigcup_{\p \not\in S} \p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$\left( \bigcap_{\p \not \in S} (\O_K \setminus \p) \right)^{-1} \O_K = \bigcap_{\p \not \in S} ((\O_K \setminus \p)^{-1} \O_K),$$
which is not obvious.
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A) We clearly have $\Sigma^{-1} \O_K \subseteq \O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $\O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x \in \O_{K,S}$. For every $\p \in S$ such that $v_{\p}(x) < 0$, write $\p^h = b_{\p} \O_K$ as an principal ideal. Then
$b := \prod\limits_{v_{\p}(x) < 0} b_{\p}^{-v_{\p}(x)}$ is an element of $\O_K$ only divisible by primes in $S$, so that $b \in \Sigma$.
Finally, $bx \in \O_K$, so we conclude that $x \in \Sigma^{-1} \O_K$ as desired.
[Here is a wrong proof that $\O_{K,S} \subset \Sigma^{-1} \O_K$ : if $x \in \O_{K,S}$ is non-zero, then fix $b_p \in p^{-v_p(x)}$ for every prime $p \in S$ such that $v_p(x)<0$. Then $b = \prod_p b_p \in \Sigma$ and $xb \in \O_K$, so that $x \in \Sigma^{-1} \O_K$. The problem is that $b$ may have prime factors outside $\{ \p \mid a_i \}$, so that $b \in \Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
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B) Once you know that $\O_{K,S} \cong \Sigma^{-1}\O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $\O_{K,S}$ is indeed an extended ideal $J \O_{K,S}$, where $J$ is an ideal of $\O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
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C) You can also look at prop. 4.19 here, but notice that they define $\O_{K,S}$ as a localization.
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2) Here we are considering the multiplicative set
$\Sigma = \O_K \setminus \bigcup\limits_{\p \in \mathrm{Spec}(\O_K) \setminus S} \p.$
If instead you consider $\Sigma' = \O_K \setminus \p$ for a single fixed prime $\p$ of $\O_K$, you also get that $(\Sigma')^{-1}\O_K = (\O_K)_{\p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $\Sigma$ deals with the cofinite set $\mathrm{Spec}(\O_K) \setminus S$ of primes, while $\Sigma'$ deals with the finite set $\{\p\}$...
Best Answer
First the geometric perspective. Let $X=\newcommand\Spec{\operatorname{Spec}}\Spec R$, $X_S=\Spec R_S$. Noting that $X_S$ is an open subset of $X$, we have a surjective map $\newcommand\Pic{\operatorname{Pic}}\Pic X\to \Pic X_S$ in the usual way. However $X_S$ is just the open subscheme of $X$ where we deleted the union of the supports of the generators of $\Pic X$, so trivially this map is the 0 map. Hence $\Pic X_S$ is 0. Thus $R_S$ is a PID, since it has trivial ideal class group.
On the other hand, if we go the classical ANT route, we have the following. (This is the same as above, but phrased in more classical language)
Since $R$ and $R_S$ have the same fraction field, any fractional ideal of $R_S$ is a fractional ideal of $R$. Moreover for any fractional ideal $\Lambda$ of $R$, $\Lambda R_S$ is a fractional ideal of $R_S$. By the first observation, this map $\Lambda\mapsto \Lambda R_S$ is surjective. Note that it is clear that this is a group homomorphism, since $\Lambda_1\Lambda_2R_S=(\Lambda_1R_S)(\Lambda_2R_S)$. Hence we have a surjective map from the group of fractional ideals of $R$ to the group of fractional ideals of $R_S$. Next we note that this map sends principal fractional ideals of $R$ to principal fractional ideals of $R_S$, since if $f\in \operatorname{Frac} R$, then $fR\mapsto fR_S$. Hence this map induces a surjective group homomorphism from the ideal class group of $R$ to the ideal class group of $R_S$.
Now we just consider what happens to $I_j$ for any of the ideals in your question. First note that since $I_j$ was an ideal of $R$, $I_jR_S$ is therefore an ideal of $R_S$ (as in not a fractional ideal). We'll show that $I_jR_S$ is not contained in any prime ideals in $R_S$, and therefore $I_jR_S$ is the unit ideal, and hence trivial in the ideal class group of $R_S$. By surjectivity of the homomorphism, this will allow us to conclude that the ideal class group of $R_S$ is in fact trivial, and that $R_S$ is therefore a PID.
Now to see that $I_jR_S$ is the unit ideal, we just observe that it isn't contained in any prime of $R_S$. This follows since the primes of $R_S$ are of the form $\newcommand\pp{\mathfrak{p}}\pp R_S$ where $\pp\not\in S$, and have the property that $\pp R_S\cap R=\pp$, so if $I_jR_S \subseteq \pp R_S$, then $I_j\subseteq I_j R_S \cap R \subseteq \pp R_S \cap R = \pp$, which is a contradiction. Hence $I_jR_S$ is the unit ideal. Thus we are done.