I am very far from giving a satisfying answer, but the fact that $S$ is saturated is immediate:
Claim. The set $S$ is saturated in the sense that if $ab \in S$ for some $a,b \in R$, then $a \in S$. Proof. Let $x \in R$ and let $I$ be the ideal of $R$ generated by $a$ and $x$. As $Ib$ is principal, so is $I$.
I am unable to produce examples of such sets $S$ which are neither $R \setminus \{0\}$ (e.g., $R$ is a Bézout domain) nor $R^{\times}$ (e.g., $R = \mathbb{Z}[X]$, a non-Bézout GCD domain). These trivial examples of $S$ are multiplicatively closed.