2 Added hypothesis to lemma; edited body

Your question is both interesting and tricky, and although I don't have an answer to the specific question you asked, I do have some observations about the situation in other Boolean algebras.

First is the trivial observation that in a Boolean algebra with atoms, the union of two incomparable regular filters certainly can generate an ultrafilter as follows: let $a$ and $b$ be two non-atomic elements whose meet $a\wedge b$ is an atom. Let $F$ be the principal filter consisting of all elements $\geq a$ and let $G$ be the principal filter of all elements $\geq b$. These are both regular in your terminology, witnessed by the corresponding principal ideals below $a$ and $b$, respectively. But the filter generated by the union $F\cup G$ will contain both $a$ and $b$ and hence $a\wedge b$, and this will be the principal ultrafilter generated by this atom.

More generally, weakening your regularity property, I observe that in any Boolean algebra there can be distinct incomparable nontrivial filters whose union generates an ultrafilter, as follows. Let $\mathbb{B}$ be any Boolean algebra with an ultrafilter $\mu$ on $\mathbb{B}$, and suppose $a\in\mu$ with $a\neq 0,1$. Let $F$ be the filter $\{x\vee \neg a\mid x\in \mu\}$ and let $G$ be any regular subfilter of $\mu$ containing $a$, such as the principal filter on $a$, consisting of all $y\geq a$. Note that $F\not\subset G\not\subset F$. Note also that $F\cup G$ has $a$ and has $x\vee \neg a$ for any $x\in\mu$, and so it has $x\wedge a=(x\vee \neg a)\wedge a$. Thus, the union filter generates $\mu$. But also, $F\cup G\subset\mu$ and so the filter it generates is contained in $\mu$, and so $F\cup G$ is contained in a unique ultrafilter, $\mu$.

In your case of $P(\omega)/\text{Fin}$, you could take $\mu$ to be any ultrafilter containing the set $a$ of even numbers, and let $F$ be all sets in $\mu$, but throwing in the odd numbers to accompany every one, and let $G$ be any subfilter of $\mu$ containing the evens, such as the principal filter on the evens. The union $F\cup G$ will contain the evens and the union of any set in $\mu$ with the odds, and so the filter it generates will contain any set in $\mu$, which is an ultrafilter.

But the filter $F$ in these examples is not regular in your sense, and so this does not answer the question.

Meanwhile, the situation is impossible in a complete atomless Boolean algebra:

Theorem. If $F$ and $G$ are regular filters in a complete atomless Boolean algebra $\mathbb{B}$, then $F\cup G$ does not generate an ultrafilter.

Proof. The main point is that if $F$ is a regular filter, as witnessed by ideal $I$, and $\mathbb{B}$ is complete, then actually $F$ and $I$ are principal. To see this, note that $F$ consists of the upper bounds of $I$, but by completeness, there is a least upper bound, and so $F$ has a minimal element $a=\text{inf}(F)=\text{sup}(I)$. Similarly, the regular filter $G$, witnessed by ideal $J$, is also principal, with $b=\text{inf}(G)=\text{sup}(J)$. It follows that $a\wedge b$ is the least element of the filter generated by $F\cup G$, and so $\langle F\cup G\rangle$ is the principal filter on $a\wedge b$. In an atomless Boolean algebra, however, principal filters are never ultrafilters. QED

The point was that regular ultrafilters in a complete Boolean algebra are the same as principal filters.

A related observation is that if $\mathbb{B}$ is not complete, then we may still complete it, forming the completion $\bar{\mathbb{B}}$, in which $\mathbb{B}$ is dense, and the idea of the previous proof shows:

Lemma. In any Boolean algebra $\mathbb{B}$, the regular filters $F$ are exactly the traces on $\mathbb{B}$ of the principal filters of the completion $\bar{\mathbb{B}}$.

Proof. If $F$ is regular in $\mathbb{B}$, as witnessed by ideal $I$, then $F$ consists precisely of the upper bounds of $I$ in $\mathbb{B}$. Let $a$ be the least upper bound of $I$ in the completion $\bar{\mathbb{B}}$, and let $\bar F$ be the principal filter on $a$ in $\bar{\mathbb{B}}$. It follows that $F\subset\bar F$, and furthermore $\bar F\cap\mathbb{B}=F$, because any element of $\mathbb{B}$ above $a$ in $\bar{\mathbb{B}}$ is an upper bound of $I$ and hence already in $F$. Thus, $F$ is the trace of the principal filter $\bar{F}$ in $\bar{\mathbb{B}}$ on $\mathbb{B}$.QED

And another observation concerns the situation arising in any nonatomic positive instance of the phenomenon:

Lemma. If $F$ and $G$ are regular filters in an atomless Boolean algebra $\mathbb{B}$, witnessed by ideals $I$ and $J$, respectively, and $F\cup G$ generates an ultrafilter, then $I$ and $J$ are orthogonal, in the sense that $i\wedge j=0$ for any $i\in I, j\in J$. Thus, the dual filter to $I$ is contained in $G$ and the dual filter to $J$ is contained in $F$.

Proof. If $i\wedge j\gt 0$ for some $i\in I$ and $j\in J$, then since $F$ consists of the upper bounds of $I$ and $G$ consists of the upper bounds of $J$, it follows that every element of $F\cup G$ is contained in the principal filter above $i\wedge j$, and since $\mathbb{B}$ is atomless this cannot be an ultrafilter. Another way to say $i\wedge j=0$ is to say that $\neg i$ is above $j$ and $\neg j$ is above $i$, and so the statement about dual filters follows. QED

I have called these both "lemmas," and they seem close to establishing something, but I don't I'm not yet quite see clearly to a sure what the conclusion is that they might be building toward.

1

Your question is both interesting and tricky, and although I don't have an answer to the specific question you asked, I do have some observations about the situation in other Boolean algebras.

First is the trivial observation that in a Boolean algebra with atoms, the union of two incomparable regular filters certainly can generate an ultrafilter as follows: let $a$ and $b$ be two non-atomic elements whose meet $a\wedge b$ is an atom. Let $F$ be the principal filter consisting of all elements $\geq a$ and let $G$ be the principal filter of all elements $\geq b$. These are both regular in your terminology, witnessed by the corresponding principal ideals below $a$ and $b$, respectively. But the filter generated by the union $F\cup G$ will contain both $a$ and $b$ and hence $a\wedge b$, and this will be the principal ultrafilter generated by this atom.

More generally, weakening your regularity property, I observe that in any Boolean algebra there can be distinct incomparable nontrivial filters whose union generates an ultrafilter, as follows. Let $\mathbb{B}$ be any Boolean algebra with an ultrafilter $\mu$ on $\mathbb{B}$, and suppose $a\in\mu$ with $a\neq 0,1$. Let $F$ be the filter $\{x\vee \neg a\mid x\in \mu\}$ and let $G$ be any regular subfilter of $\mu$ containing $a$, such as the principal filter on $a$, consisting of all $y\geq a$. Note that $F\not\subset G\not\subset F$. Note also that $F\cup G$ has $a$ and has $x\vee \neg a$ for any $x\in\mu$, and so it has $x\wedge a=(x\vee \neg a)\wedge a$. Thus, the union filter generates $\mu$. But also, $F\cup G\subset\mu$ and so the filter it generates is contained in $\mu$, and so $F\cup G$ is contained in a unique ultrafilter, $\mu$.

In your case of $P(\omega)/\text{Fin}$, you could take $\mu$ to be any ultrafilter containing the set $a$ of even numbers, and let $F$ be all sets in $\mu$, but throwing in the odd numbers to accompany every one, and let $G$ be any subfilter of $\mu$ containing the evens, such as the principal filter on the evens. The union $F\cup G$ will contain the evens and the union of any set in $\mu$ with the odds, and so the filter it generates will contain any set in $\mu$, which is an ultrafilter.

But the filter $F$ in these examples is not regular in your sense, and so this does not answer the question.

Meanwhile, the situation is impossible in a complete atomless Boolean algebra:

Theorem. If $F$ and $G$ are regular filters in a complete atomless Boolean algebra $\mathbb{B}$, then $F\cup G$ does not generate an ultrafilter.

Proof. The main point is that if $F$ is a regular filter, as witnessed by ideal $I$, and $\mathbb{B}$ is complete, then actually $F$ and $I$ are principal. To see this, note that $F$ consists of the upper bounds of $I$, but by completeness, there is a least upper bound, and so $F$ has a minimal element $a=\text{inf}(F)=\text{sup}(I)$. Similarly, the regular filter $G$, witnessed by ideal $J$, is also principal, with $b=\text{inf}(G)=\text{sup}(J)$. It follows that $a\wedge b$ is the least element of the filter generated by $F\cup G$, and so $\langle F\cup G\rangle$ is the principal filter on $a\wedge b$. In an atomless Boolean algebra, however, principal filters are never ultrafilters. QED

The point was that regular ultrafilters in a complete Boolean algebra are the same as principal filters.

A related observation is that if $\mathbb{B}$ is not complete, then we may still complete it, forming the completion $\bar{\mathbb{B}}$, in which $\mathbb{B}$ is dense, and the idea of the previous proof shows:

Lemma. In any Boolean algebra $\mathbb{B}$, the regular filters $F$ are exactly the traces on $\mathbb{B}$ of the principal filters of the completion $\bar{\mathbb{B}}$.

Proof. If $F$ is regular in $\mathbb{B}$, as witnessed by ideal $I$, then $F$ consists precisely of the upper bounds of $I$ in $\mathbb{B}$. Let $a$ be the least upper bound of $I$ in the completion $\bar{\mathbb{B}}$, and let $\bar F$ be the principal filter on $a$ in $\bar{\mathbb{B}}$. It follows that $F\subset\bar F$, and furthermore $\bar F\cap\mathbb{B}=F$, because any element of $\mathbb{B}$ above $a$ in $\bar{\mathbb{B}}$ is an upper bound of $I$ and hence already in $F$. Thus, $F$ is the trace of the principal filter $\bar{F}$ in $\bar{\mathbb{B}}$ on $\mathbb{B}$.QED

And another observation concerns the situation arising in any nonatomic positive instance of the phenomenon:

Lemma. If $F$ and $G$ are regular filters in an atomless Boolean algebra $\mathbb{B}$, witnessed by ideals $I$ and $J$, respectively, then $I$ and $J$ are orthogonal, in the sense that $i\wedge j=0$ for any $i\in I, j\in J$. Thus, the dual filter to $I$ is contained in $G$ and the dual filter to $J$ is contained in $F$.

Proof. If $i\wedge j\gt 0$ for some $i\in I$ and $j\in J$, then since $F$ consists of the upper bounds of $I$ and $G$ consists of the upper bounds of $J$, it follows that every element of $F\cup G$ is contained in the principal filter above $i\wedge j$, and since $\mathbb{B}$ is atomless this cannot be an ultrafilter. Another way to say $i\wedge j=0$ is to say that $\neg i$ is above $j$ and $\neg j$ is above $i$, and so the statement about dual filters follows. QED

I have called these both "lemmas," and they seem close to establishing something, but I don't yet quite see clearly to a conclusion.