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This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated.

Let us denote by $D^{\times}$ the unit group of $D$. We have $i(D) = D^{\times}$ for $D = k[x, y, z]$. Indeed, Krull's Principal Ideal Theorem shows that $i(D) = D^{\times}$ holds for any Noetherian domain whose maximal ideals have height at least $3$.

The two trivial sets $D \setminus \{0\}$ (e.g., $D$ is a Dedekind domain) and $D^{\times}$ (e.g., $D$ as above) are clearly multiplicatively closed and saturated.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

This yields almost instantly

Claim 1. Let $a, b \in i(D)$ be such that $Da + Db = R$. Then $ab\in i(D)$.

Proof. We have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$.

In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$.

Now comes the good part

Claim 2. The set $i(D)$ is saturated.

Proof. Let $a,b \in D$ be such that $ab \in i(D)$. Then $D/Dab$ is a PIR, and so is its quotient $D/Da$. Therefore $a \in i(D)$.

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references.

This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated.

Let us denote by $D^{\times}$ the unit group of $D$. We have $i(D) = D^{\times}$ for $D = k[x, y, z]$. Indeed, Krull's Principal Ideal Theorem shows that $i(D) = D^{\times}$ holds for any Noetherian domain whose maximal ideals have height at least $3$.

The two trivial sets $D \setminus \{0\}$ (e.g., $D$ is a Dedekind domain) and $D^{\times}$ (e.g., $D$ as above) are clearly multiplicatively closed and saturated.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

This yields almost instantly

Claim 1. Let $a, b \in i(D)$ be such that $Da + Db = R$. Then $ab\in i(D)$.

Proof. We have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$.

In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$.

Now comes the good part

Claim 2. The set $i(D)$ is saturated.

Proof. Let $a,b \in D$ be such that $ab \in i(D)$. Then $D/Dab$ is a PIR, and so is its quotient $D/Da$. Therefore $a \in i(D)$.

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references. The condition $i(D) = D \setminus \{0\}$ implies that any ideal of $D$ is generated by two elements, i.e., $D$ is a Bass domain. The class of Bass rings has a well-developed theory, see Bass’s Work in Ring Theory and Projective Modules by T. Y. Lam.

This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated.

We denote by $D^{\times}$ the unit group of $D$.

We clearly have $i(D) = D \setminus \{ 0 \}$ for any Bézout domain $D$, while $i(D) = D^{\times}$ for $D = k[x, y, z]$. Krull's Principal Ideal Theorem shows that $i(D) = D^{\times}$ holds for any Noetherian domain whose maximal ideals have height at least $3$.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

This yields almost instantly

Claim 1. Let $a, b \in i(D)$ be such that $Da + Db = R$. Then $ab\in i(D)$.

Proof. We have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$.

In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$.

Now comes the good part

Claim 2. The set $i(D)$ is saturated.

Proof. Let $a,b \in D$ be such that $ab \in i(D)$. Then $D/Dab$ is a PIR, and so is its quotient $D/Da$. Therefore $a \in i(D)$.

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references.

This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated.

Let us denote by $D^{\times}$ the unit group of $D$. We have $i(D) = D^{\times}$ for $D = k[x, y, z]$. Indeed, Krull's Principal Ideal Theorem shows that $i(D) = D^{\times}$ holds for any Noetherian domain whose maximal ideals have height at least $3$.

The two trivial sets $D \setminus \{0\}$ (e.g., $D$ is a Dedekind domain) and $D^{\times}$ (e.g., $D$ as above) are clearly multiplicatively closed and saturated.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

This yields almost instantly

Claim 1. Let $a, b \in i(D)$ be such that $Da + Db = R$. Then $ab\in i(D)$.

Proof. We have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$.

In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$.

Now comes the good part

Claim 2. The set $i(D)$ is saturated.

Proof. Let $a,b \in D$ be such that $ab \in i(D)$. Then $D/Dab$ is a PIR, and so is its quotient $D/Da$. Therefore $a \in i(D)$.

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references.

This is only a long comment.

We clearly have $i(D) = D \setminus \{ 0 \}$ for any Bézout domain $D$, while $i(k[x, y, z]) = k \setminus \{ 0 \}$ by Krull's principal ideal theorem.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

If $a, b \in i(D)$ are coprime, i.e., $Da + Db = R$, then we have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$. In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$. (The set $i(D)$ is trivially saturated).

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references.

This is only half of an answer. Claim 2 below will establish that $i(D)$ is always saturated.

We denote by $D^{\times}$ the unit group of $D$.

We clearly have $i(D) = D \setminus \{ 0 \}$ for any Bézout domain $D$, while $i(D) = D^{\times}$ for $D = k[x, y, z]$. Krull's Principal Ideal Theorem shows that $i(D) = D^{\times}$ holds for any Noetherian domain whose maximal ideals have height at least $3$.

As exemplified by js21 with $D = k[x, y]$ in a comment above, the answer to OP's first question is no in general. Still, I'll provide a small glint of hope.

OP's definition can be rephrased as follows.

Rewording. The set $i(D)$ consists of the elements $a \in D \setminus \{ 0 \}$ such that $D/Da$ is a principal ideal ring (PIR).

This yields almost instantly

Claim 1. Let $a, b \in i(D)$ be such that $Da + Db = R$. Then $ab\in i(D)$.

Proof. We have $D/Dab \simeq D/Da \times D/Db$ by the Chinese Remainder Theorem. Since a direct product two PIRs is a PIR, we infer that $ab \in i(D)$.

In particular the set of primes $p \in D$ such that $Dp$ is a maximal ideal of $D$ is a submonoid of $D \setminus \{0\}$ contained in $i(D)$.

Using the above rewording, we can quickly check why js21's comment answers the question in the negative. If $k$ is field, we have $k[x, y]/(x) \simeq k[y]$ which is a principal ideal domain whereas $k[x, y]/(x^2) \simeq (k[x]/(x^2))[y]$ is not a PIR (the ideal generated the images of $x$ and $y$ is readily seen not to be principal). Therefore, we have $x \in i(D)$ whereas $x^2 \notin i(D)$.

Let us give another example for which multiplicative closedness of $i(D)$ fails dramatically. Consider a regular local ring $D$ (this is necessarily a domain) of Krull dimension $2$ with maximal ideal $\mathfrak{m}$. Then $i(D) = (D \setminus \mathfrak{m}) \cup (\mathfrak{m} \setminus \mathfrak{m}^2)$, but no product of two non-units in $i(D)$ belongs to $i(D)$.

Now comes the good part

Claim 2. The set $i(D)$ is saturated.

Proof. Let $a,b \in D$ be such that $ab \in i(D)$. Then $D/Dab$ is a PIR, and so is its quotient $D/Da$. Therefore $a \in i(D)$.

Side note. The answers to this MSE question show why $i(D) = D \setminus \{0\}$ when $D$ is a Dedekind domain and point to interesting references.