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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. 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 $R$.

The answer is yes is $R$ is any Noetherian domain.

For every integral domain $R$, the set $S = S_R$ is saturated:

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

Proof. Let $x \in R$ and let $I$ be the ideal generated by $a$ and $x$. 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 a Noetherian domain, then $S_R$ is multiplicatively closed.

Proof. If $S_R$ contains no irreducible element, then $S_R = R^{\times}$ because $S_R$ is saturated and any element of $R$ is a product of finitely many irreducible elements. If $S_R$ contains a single irreducible element $p$, it suffices to show that it contains any power of $p$. Let us suppose that $p^k \in S_R$ for some $k \ge 1$ and let $x \in R$. Since $Rp^k + Rx = Rd$ for some $d \in R$, we have $d \vert p^k$ and hence $d \in S_R$ for $S_R$ is saturated. Therefore $p \vert d$, so that $p \vert x$. we write no $x = px'$ with $e \in R$. As $Rp^k + Rx' = Rd'$ for some $d' \in R$, we obtain $Rp^{k + 1} + Rx = Rd'p$ by multiplying the left and right hand sides by $p$. The result follows by induction on $k$. Assume now there are two distinct irreducible elements $p, q \in R$ in $S_R$. Given $x \in R$, there is some $d \in R$ such that $Rp + Rq = Rd$. As $d$ divides both $p$ and $q$, it is a unit. Thus $Rp + Rq = R$. Let $x \in R$. If $x \in Rp \cup Rq$, it is easily checked that $Rpq + Rx$ is principal by dividing this ideal by $p$ or $q$. Otherwise $R = Rp + Rx = Rq + Rx$. Thus $1$ belongs to $Rpq + Rx$, i.e., $Rpq + Rx = R$.

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 unique factorization domain).

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 $R$, 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 $S_{R[X]} = (R[X])^{\times}$.

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 $Rp$ is the unique maximal ideal of $R$.

Proof. Let $x \in R$ and let $I$ be the ideal of $R$ generated by $a$ and $x$. As $Ib$ is principal, so is $I$.

OP's first claim. We have $S_R = N_R$ if $R$ is a local domain.

OP's second claim. The set $N_R$ is multiplicatively closed and saturated for every integral domain $R$.

Corollary. If $R$ is a regular local domain which is not principal, then $S_R = N_R = R^{\times}.$

Proof. As $R$ is a UFD with at least two prime elements, the result follows from the identity $S_R = N_R$.

Proof. Let $x \in R$ and let $I$ be the ideal generated by $a$ and $x$. As $Ib$ is principal, so is $I$.

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.

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 $R$.

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 UFD). 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 $r \in R$.

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 $\mathbb{Z}_{\mathcal{P}}$.

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 $r \in R$.

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. We have $S_R = N_R$ if $R$ is a local domain.

OP's second claim. The set $N_R$ is multiplicatively closed and saturated for every integral domain $R$.

From this, we infer

Corollary. If $R$ is a regular local domain which is not principal, then $S_R = N_R = R^{\times}.$

Proof. As $R$ is a UFD with at least two prime elements, the result follows from the identity $S_R = N_R$.