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 generated by $a$ and $x$. As $Ib$ is principal, so is $I$. 

Let us denote by $R^{\times}$ the unit group of $R$.
I can only produce examples of sets $S_R$ which are either $R \setminus \{0\}$ (obtained, e.g., when $R$ is a Bézout domain) or $R^{\times}$ (obtained, e.g., for $R = \mathbb{Z}[X]$, a non-Bézout UFD). These trivial examples of $S_R$ are certainly 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 $S = S_{R[X]}$ 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) with infinitely many primes or that there is no non-constant $u$-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 $a \in R \setminus \{0\}$. 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-zero element $a$ of $R$. If $R$ is a UFD, 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 over UFDs which aren't fields **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 $\mathbb{Z}_{\mathcal{P}}$.

The OP defines a seemingly narrower set $N_R = \{ a \in R \setminus \{0\}\} : a \vert x \text{ or } x \vert a \text{ for every } x \in R\}$ which satisfies $$R^{\times} \subset N_R \subset S_R \subset R \setminus \{0\}.$$


>> **OP's first claim**. If $R$ is a local domain, then $S_R = N_R$.  

>> **OP's second claim**. If $R$ is any integral domain, then the set $N_R$ is multiplicatively closed and saturated.

From this, we infer

>>**Corollary 1.** If $R$ is a non-Bézout UFD, then $N_R = R^{\times}.$

>>*Proof.* As $R$ has at least two distinct prime elements, $N_R$ cannot contain any prime element. Since $N_R$ is saturated, it cannot contain any non-unit element.

In particular, we have $S_R = R^{\times}$ for any regular local domain $R$.

We obtain an alternative for Noetherian local domains $R$ with Krull dimension $1$: either $R$ is a principal ideal domain and obviously $S_R = R \setminus \{ 0 \}$, or $R$ isn't and $S_R = R^{\times}$. This follows from

>>**Corollary 2.**. Assume that $R$ is a Noetherian domain such that $S_R = N_R$, e.g, $R$ is a Noetherian local domain. Then either $R$ is a local domain with a principal maximal ideal and $S_R = R \setminus \{0\}$ or $S_R = R^{\times}$.

>>*Proof.* Assume that $S_R$ contains a non-unit element. As $R$ is Noetherian and $S_R$ is saturated, the set $S_R$ contains an irreducible element $p$. Since $p \in N_R$, the element $p$ divides any non-unit of $R$. Therefore $(p)$ is the unique maximal ideal of $R$.

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[1] S. H. Weintraub, "Values of polynomials over integral domains", 2014.