Claim:
$ int(A\cup B) = int(A) \cup int(B) \iff \partial A \cap \partial B \subset \partial (A \cup B) $
Proof:
Since you are more intrested in the ($\Leftarrow$) part of the claim I will first prove that. For both parts of the proof, remeber that interior, exterior and boundary of a set is a partitioning of the whole space i.e. $int(A) \cup \partial A \cup ext(A) = X$ where all three sets are disjoint. Also, interior and exterior are open sets.
($\Leftarrow$)
As you have said $int(A) \cup int(B) \subset int(A\cup B)$. So, I will only focus on proving $int(A\cup B) \subset int(A) \cup int(B)$.
Let $x \in int (A \cup B)$. If $x \in int(A)$ or $x \in int(B)$ then we are done. So, assume that $x \notin int(A)$ and $x \notin int(B)$.
By hypothesis, if $x \in \partial A \cap \partial B$ then $x \in \partial(A\cup B)$ but interior and boundary of $(A \cup B)$ are disjoint which is a contradiction. So, $x \notin \partial A \cap \partial B$.
Without loss of generality, assume that $x \notin \partial B$. Then, $x \in ext(B)$ and either $x \in \partial A$ or $x \in ext(A)$.
Case 1: $x \in \partial A$,
Since $x \in ext(B)$, there is a neighborhood $V \subset B^C$ of x. Since $x \in int(A \cup B)$ there is a neighborhood $U \subset (A\cup B)$ of x. Then,
$U \cap V$ is a neighborhood of $x$ in $X$. Note that $(U \cap V) \subset (A\cup B) \cap B^C \subset A$. But since $x \in \partial A$, any neighborhood of $x$ must intersect non-trivially with $A^C$. Hence, it is a contradiction. So, $x \notin \partial A$.
Case 2: $x \in ext(A)$,
$x \in int(A \cup B) \subset A \cup B \subset \bar A \cup \bar B = (int(A) \cup
\partial A) \cup (int(B) \cup \partial B)$. However $x \in ext(A)$ and $x \in ext(B)$ implying that $x \notin int(A) \cup \partial A$ and
$x \notin int(B) \cup \partial B$ which is a contradiction.
Therefore, $x \in int(A)$ or $x \in int(B)$ i.e. $x \in int(A) \cup int (B)$.
$ (\Rightarrow) $ Let $x \in \partial A \cap \partial B$.
So, $x \notin int(A) $ and $ x \notin int(B)$ implying $x \notin int(A) \cup int(B) = int(A\cup B)$.
Thus, $x \in \partial (A \cup B) \cup ext(A\cup B)$.
Assume $x \in ext(A \cup B)$ then there is a neighborhood $P \subset (A \cup B)^C $ of $x$ (where $A^C$ means the complement for set $A$).
But $(A \cup B)^C \subset A^C$. That is, we have a neighborhood of $x$ which is contained in $A^C$ meaning $ x \notin \partial A$ which is a contradiction.
Hence, $x \in \partial(A\cup B)$.
This is true if and only if the boundary has empty interior.
It's easy to see the boundary of the boundary is always contained in the boundary. But if the boundary has an interior point, then that point will not be in the boundary of the boundary since it is not in the closure of the complement.
Edit: to answer your added question, it is always true that $bbbA=bbA$. This follows from the above and the fact that $ibcB$ is empty for any $B$ and the fact that the boundary of a set is closed, so letting $B=bA$, so $cB=B$, and $ibbA$ is empty. To see that $ibcB$ is empty, assume not and let $x$ be an interior point of the boundary of a closed set $C$. Then the boundary contains an open set $U$ containing $x$. Since $C$ is closed it contains its boundary so it contains $U$. So $U$ does not intersect $kC$, so $x$ is not in $ckC$, so it is not in $bC$, contradiction.
Best Answer
Let it be that $\overline{A}\cap B=\emptyset=A\cap\overline{B}$.
If $x\in\text{int}\left(A\cup B\right)$ and $x\in A$ then also $x\in\text{int}\left(A\cup B\right)\cap\overline{B}^{c}\subseteq A$.
Note that $\text{int}\left(A\cup B\right)\cap\overline{B}^{c}$ is open so we are allowed to conclude that $x\in\text{int}\left(A\right)$.
If $x\in\text{int}\left(A\cup B\right)$ and $x\in B$ then likewise we find $x\in\text{int}\left(B\right)$
Proved is now that $\text{int}\left(A\cup B\right)\subseteq\text{int}A\cup\text{int}B$ and it is obvious that also $\text{int}A\cup\text{int}B\subseteq\text{int}\left(A\cup B\right)$.