I am very far from giving a satisfying answer, but the fact that $S$ is saturated is immediate:
Claim 1. The set $S = S_R$ 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 $I$.
Let us denote by $R^{\times}$ the unit group of $R$. I am unable to produce examples of sets $S_R$ which are neither $R \setminus \{0\}$ (obtained, e.g., when $R$ is a Bézout domain) nor $R^{\times}$ (obtained, e.g., for $R = \mathbb{Z}[X]$, a non-Bézout GCD domain). These trivial examples of $S_R$ are multiplicatively closed.
Still, Claim 1 can be used to inspect the state of affairs regarding polynomial rings $R[X]$ with $R$ an integral domain. For those rings, one could dream of an alternative. Indeed, $S = S_{R[X]}$ is one of the two trivial sets on many instances. We call $f \in R[X]$ a $u$-polynomial if $f(r)$ is a unit of $R$ for every $r \in R$.
Claim 2. Let $R$ be an integral domain which is not a field. Assume moreover that $R$ is a unique factorization domain (UFD) or that there is no $u$-polynomial over $R$. Then $S_{R[X]} = (R[X])^{\times}$.
Proof. We begin with an observation. Let $a, b \in R$. As the ideal generated by $a$ and $X - b$ is principal if and only if $a \in R^{\times}$, we deduce that $a \notin S_{R[X]}$ and $X - b \notin S_{R[X]}$ for every $a \in R \setminus R^{\times}$ and every $b \in R$. Now let $f \in S_{R[X]}$ and let $a \in R \setminus R^{\times}$. The ideal generated by $f$ and $a$ is a principal ideal generated by some $g \in R[X]$. Since $g$ divides $a$ and $f$, it is a constant polynomial which lies in $S_{R[X]}$ by Claim 1. Hence $g$ is unit of $R$ by the above remark. As result, the reduction of $f$ modulo $Ra$ is a unit of $(R/aR)[X]$ for every non-unit $a$ of $R$. If $R$ is a GCD, then $f$ must be a constant polynomial, hence a unit. Otherwise, let us consider the ideal generated by $f$ and $X - b$ for some $b \in R$. It is a principal ideal generated by some $h \in R[X]$ which divides both $f$ and $X - b$. As it cannot be $X - b$ multiplied by a unit of $R$ by our first remark, it is a unit of $R$. Therefore $f(b)$ is a unit of $R$ too. Since this holds for every $b \in R$, $f$ is $u$-polynomial.
Note that non-constant $u$-polynomials do exist, see e.g., [1, Example 3.b]. Indeed, take $\mathcal{P} =\{p \text{ prime } \vert\, p \equiv 3 \text{ mod } 4 \} \subset \mathbb{Z}$ and set $$\mathbb{Z}_{\mathcal{P}} = \{ \text{ rational numbers } \frac{m}{n} \text{ with no prime factor of } n \text{ in } \mathcal{P}\}.$$ Then $x^2 + 1$ is $u$-polynomial over the UFD $\mathbb{Z}_{\mathcal{P}}$.
[1] S. H. Weintraub, "Values of polynomials over integral domains", 2014.