This answer addresses the trivial part of the question, that is multiplicative stability, while providing **non-trivial examples of saturated sets $k(R)$** under two rather restrictive conditions: $R$ is a Dedekind domain or a polynomial ring over a unique factorization domain (UFD).

It is easy to show that $k(R)$ is multiplicatively closed.

**Claim.** Let $R$ be an integral domain. Then $k(R)$ is multiplicatively closed.

*Proof.* Let $x, y \in k(R)$ and let $b \in R$. If $Rx + Rb = Ry + Rb = R$, then $Rxy + Rb = R$ since $1 \in (Rx + Rb)(Ry + Rb) \subset Rxy + R$. Otherwise, there is a non-invertible element $d \in R$ such that $Rx + Rb \subset Rd$ or $Ry + Rb \subset Rd$. In both cases, we have $Rxy + Rb \subset Rd$, which completes the proof.

I **fail** to find an integral domain $R$ such that **$k(R)$ is not saturated**.
The two propositions below provide conditions under which $k(R)$ is saturated, yielding **non-trivial examples** of sets $k(R)$.
By trivial, I mean one of the two possibilities $k(R) = R^{\times}$ (the unit group of $R$) or $k(R) = R \setminus \{0\}$. The latter is the subject of a theorem of Robert Gilmer and William Heinzer [1, Corollary 2]:

**Gilmer-Heinzer's Theorem.** Let $R$ be an integral domain. If $k(R) = R \setminus \{0\}$ and $R$ satisfies the ascending chain condition on principal ideals (ACCP), then $R$ is a principal ideal ring (PID).

The authors define the *Kummer condition*, or condition $(K)$, as $k(R) = R \setminus \{0\}$. According to the references of [1], Condition $(K)$ is at the core of Ernst Kummer's error in his proof of that Fermat's last theorem holds for regular primes. Kummer would have assumed that $k(R) = R \setminus \{0\}$ holds for $R = \mathbb{Z}[\zeta_n]$ with $\zeta_n = e^{\frac{2i\pi}{n}}$ or $R = \mathbb{Z}[\zeta_n][X]$. It is nowadays well-known that any of the two previous Noetherian domains can fail to be a UFD, e.g., for $n = 23$.

The following remarks will come soon in handy.

**Remark 1.** Let $R$ be an integral domain. Then $a \in k(R)$ if and only if $Ra + Rb = R$ (in other words, *$a$ is coprime with $b$*) whenever $a$ and $b$ have no non-invertible common divisor (in other words, $\text{gcd}(a, b) = 1$).

**Remark 2.** Let $R$ be an integral domain. Then $k(R)$ contains
the multiplicative submonoid $M(R)$ of $R \setminus \{0\}$ generated by the units of $R$ and the prime elements $p \in R$ such that $Rp$ is a maximal ideal of $R$. The monoid $M(R)$ is saturated in the sense that, for every $x, y \in R$, $xy \in M(R)$ implies $x \in M(R)$.

**Remark 3.** Let $R$ be an atomic domain. If $k(R)$ is saturated, then $k(R) = M(R)$.

**Remark 4.** Let $R$ be an integral domain. Let $p$ be a prime element of $R$ and let $n > 0$ such that $p^n \in R$. Then $p \in R$.

We establish first that $k(R)$ is saturated for $R$ an arbitrary Dedekind domain.

**Proposition 1.** Let $R$ be a Dedekind domain. Then $k(R) = M(R)$. In particular, $k(R)$ is saturated.

*Proof.* Given $a \in R \setminus \{0\}$, we can write
$Ra = \mathfrak{m}_1^{k_1}\mathfrak{m}_2^{k_2} \cdots \mathfrak{m}_n^{k_n}$, where each $\mathfrak{m_i}$ is a maximal ideal of $R$ and $k_i \ge 1$ an integer. Let us assume that one of the ideals $\mathfrak{m_i}$, say $\mathfrak{m_1}$, is not principal. Then we can find $b \in R$ such that $\mathfrak{m_1} = Ra + Rb$ (see, e.g., this MSE post). The elements $a$ and $b$ don't have any common non-invertible divisor, since otherwise $\mathfrak{m_1}$ wouldn't be maximal. Therefore $a \notin k(R)$. Apply now Remarks 1 and 2 to complete the proof.

The class of polynomial rings over UFDs provide other examples of sets $k(R)$ which are saturated.

**Proposition 2.** Let $S$ be a principal ideal domain (PID) and set $R \Doteq S[X]$. Then $k(R)$ is saturated. More precisely, we have

$k(R) = R^{\times}$ if $S$ has infinitely many prime elements up to multiplication by units.

$k(R) = R \setminus \{0\}$ if $S$ is a field.

$k(R) = Rp_1 \cdots p_n + R^{\times}$, if $S$ has $n \ge 1$ primes elements up to multiplication by units.

*Proof*.
Let us assume first that $S$ has infinitely many non-associated prime elements. Let $P(X) \in k(R)$. Since $S$ is a PID, there are infinitely many non-associated prime elements of $S$ which doesn't divide $P(X)$. Let $p$ be one of them. By hypothesis, $RP(X) + Rp = R$. Thus the reduction of $P(X)$ modulo $p$ is a unit of $R/pR = (S/Sp)[X]$. From this we infer that $p$ divides every coefficient of $P(X)$ except the coefficient of degree zero. Since this holds for infinitely many $p$, the polynomial $P(X)$ is constant. As $RP(X) + RX = R$ must hold, we deduce that $P(X) \in S^{\times} = R^{\times}$. Conversely, any unit of $R$ belongs to $k(R)$. Thus $k(R) = R^{\times}$.

Let us assume now that $S$ has only finitely many non-associated prime elements and let $p$ be one of them. Given $P(X) \in k(R)$, we can find $s \in S$ such that $X - s$ doesn't divide $P(X)$, so that $RP(X) + R(X - s) = R$. Therefore $P(s)$ is a unit of $S$ and hence $p$ cannot divide $P(X)$. From this we infer that $RP(X) + Rp = R$, i.e., $P(X) \in Rp + R^{\times}$. Since this holds for any prime $p$ of $S$, we deduce from the Chinese Remainder Theorem that $P(X) \in R \pi + R^{\times}$, where $\pi$ is the product of $n$ non-associated prime elements of $S$.
We shall establish the converse statement, i.e., that an arbitrary element $P(X)$ of $R \pi + R^{\times}$ lies in $k(R)$. Let $Q(X) \in R$. If $P(X)$ and $Q(X)$ have a common non-invertible divisor, then $RP(X) + RQ(X)$ is contained in a proper ideal of $R$. Thus we can assume that the greatest common divisor (gdc) of $P(X)$ and $Q(X)$ is $1$. Since the image of $\pi$ in $A \Doteq R/RP(X)$ is a unit, the ring $A$ is the quotient of $F/FP(X)$ where $F = \text{Frac}(S)[X]$ and $\text{Frac}(S)$ is the fraction field of $S$. As $\text{gcd}(P(X), Q(X)) = 1$, we have $FP(X) + FQ(X) = F$, i.e., the image of $Q(X)$ in $A$ is a unit. Equivalently, $RP(X) + RQ(X) = R$ as desired.

In order to check that $k(R)$ is saturated in the latter case, consider $P(X), Q(X) \in R$ such that $P(X)Q(X) \in k(R)$. If $P(X)$ doesn't lie in $k(R)$, then its reduction modulo $p$ for some prime $p$ of $S$ is not a unit of $(S/pS)[X]$. As the reduction of the of $P(X)Q(X)$ is a unit, we obtain a contradiction. Hence $P(X) \in k(R)$.

Eventually, if $S$ has no prime element, i.e, $S$ is a field, then $R$ is a principal ideal domain. The result follows.

The above proof shows actually a bit more than what was stated:

**Statement.** Let $S$ be a unique factorization domain (UFD) and set $R \Doteq S[X]$. Then $k(R)$ is the set of polynomials $P(X) \in R$ such that the reduction of $P(X)$ modulo $p$ is a unit of $(S/Sp)[X]$ for every prime element $p$ of $S$. In particular $k(R)$ is saturated.

**Remark 5.** Let $R$ be a UFD. Then $x \in k(R)$ if and only if $x$ is congruent either to zero or to a unit modulo $Rp$ for every prime element $p \in R$.

**Remark 6.** Let $R$ be a UFD. Let $x$ and $y$ be two distinct primes of $R$. If $xy \in k(R)$, then $Rx + Ry = R$.

[1] R. Gilmer, W. Heinzer, "Principal ideal rings and a condition of Kummer", 1982.