The answer is **yes** if $R$ is any atomic domain, e.g., $R$ is a Noetherian domain.

**Claim 1.** Let $R$ be any integral domain. The set $S = S_R$ is saturated in the sense that if $ab \in S_R$ for some $a,b \in R$, then $a \in S_R$.

*Proof.* Let $x \in R$ and let $I = Ra + Rx$. As $Ib$ is principal, so is $I$.

**Lemma.** Let $R$ be any integral domain. If $p \in S_R$ is irreducible, then $Rp$ is maximal. In particular, $p$ is prime.

*Proof.* Let $p$ be an irreducible element in $S_R$. Given $x \in R \setminus Rp$, there is $d \in R$ such that $Rp + Rx = Rd$. As $d$ divides both $p$ and $x$ and $p$ doesn't divide $x$, the element $d$ is a unit. Therefore $Rp + Rx = R$.

**Claim 2.** If $R$ is an atomic domain, then $S_R$ is multiplicatively closed.

*Proof.* Since $S_R$ is saturated and $R$ is atomic, it suffices to show that $pa \in S_R$ for every irreducible element $p \in S_R$ and every $a \in S_R$.
To do so, consider $x \in R$. If $x \in Rp$, write $x = px'$. As $a \in S_R$, there is $d \in R$ such that $Ra + Rx' = Rd$. Hence $Rpa + Rx = Rpd$. If $x \notin Rp$, then $Rp + Rx = R$ by the above lemma. As $a \in S_R$, we can find $d \in R$ such that $Ra + Rx = Rd$. Writing $1$ (resp. $d$) as a linear combination of $p$ and $x$ (resp. of $a$ and $x$), we observe that $d = 1 \cdot d \in Rpa + Rx$. Therefore $Rpa + Rx = Rd$, which completes the proof.

Let us denote by $R^{\times}$ the unit group of $R$.
The obvious examples of sets $S_R$ are $R \setminus \{0\}$ (obtained, e.g., when $R$ is a Bézout domain) and $R^{\times}$ (obtained, e.g., for $R = \mathbb{Z}[X]$, a non-Bézout unique factorization domain).

If $R$ is atomic, the above proof shows that $S_R$ is the multiplicative submonoid of $R \setminus \{0\}$ generated the units of $R$ and the prime elements $p$ of $R$ such that $Rp$ is a maximal ideal of $R$. Hence it is not too difficult to find rings $R$ such that $R^{\times} \subsetneq S_R \subsetneq R \setminus \{0\}$. Indeed, we have

**Corollary.** Let $R$ be a Dedekind domain which is not a PID, e.g., $R$ is the ring of integers of a number field with class number $> 1$. Then $S_R \subsetneq R \setminus \{0\}$.

*Proof.* By assumption, we can find a maximal ideal $\mathfrak{m}$ which is not principal. Let $a \in \mathfrak{m} \setminus \{0\}$. The element $a$ cannot be a product of prime elements since otherwise $\mathfrak{m}$ would be principal ideal generated by one of them. As a result, $a \notin S_R$.

In the case $R = \mathbb{Z}[\sqrt{-5}]$, we have $S_R = R \setminus \mathfrak{m}$ where $\mathfrak{m} = R \cdot 2 + R \cdot (1 + \sqrt{-5})$. More generally, if $R$ is the ring of integers of some number field, then $S_R$ is the complement of the union of finitely many non-principal maximal ideals.

Claim 1 can be used to inspect the state of affairs regarding polynomial rings $R[X]$ with $R$ an integral domain. For those rings $R$, the set $S = S_{R[X]}$ 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 3.** 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 $S_{R[X]} = (R[X])^{\times} \simeq R^{\times}$.

*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 with infinitely many primes, 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 $\mathbb{Z}_{\mathcal{P}}$.

In the course of Claim 3's proof, we showed

**Claim 4.** Let $R$ be an integral domain which is not a field. Then
every element of $S_{R[X]}$ is a $u$-polynomial over $R$.

This easily yields

**Claim 5.** Let $R$ be any integral domain. Then
$S_{R[X, Y]} = (R[X, Y])^{\times} \simeq R^{\times}$.

*Proof.* Cleary, $R[X]$ is not a field. Let $f(X, Y) \in (R[X])[Y]$ be a non-constant polynomial with coefficients in $R[X]$. Then $f(X, X^n)$ is a non-constant polynomial with coefficients in $R$ for all $n$ large enough. Therefore $f(X, Y)$ isn't a $u$-polynomial over $R[X]$. Consequently, $u$-polynomials over $R[X]$ are constant polynomials and the result follows from Claim 4.

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 $N_R$ is multiplicatively closed and saturated.

From this, we infer

**Corollary 1.** If $R$ is an atomic domain with at least two non-associated irreducible elements, then $N_R = R^{\times}.$

*Proof.* As $R$ has at least two non-associated irreducible elements, $N_R$ cannot contain any irreducible element. Since $N_R$ is saturated and $R$ is atomic, $N_R$ cannot contain any non-unit element.

In particular, we have $S_R = R^{\times}$ for any regular local domain $R$ which is not a principal ideal domain. We also obtain

**Remark.** If $R$ is a principal ideal domain with more than one prime, then $R^{\times} = N_R \subsetneq S_R = R \setminus \{ 0 \}$.

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 an atomic domain such that $S_R = N_R$, e.g, $R$ is a Noetherian local domain. Then either $R$ is a discrete valuation ring (DVR) and $S_R = R \setminus \{0\}$ or $S_R = R^{\times}$.

*Proof.* Assume that $S_R$ contains a non-unit element. As $R$ is atomic 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 $Rp$ is the unique maximal ideal of $R$. Since an irreducible element of $R$ lies necessarily in $Rp$, $p$ is the unique irreducible element of $R$, up to multiplication by a unit. Since $R$ is atomic $R$, we deduce that $R$ is a DVR.

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