I am wondering about a question which I have to assume has been studied, either for its own sake or sporadically in the context of other questions, but I've never come across any papers on the matter. I'm going to ask about a very simplified case.
Suppose we are given two pairs $(U, V)$ and $(Y, Z)$ of pairwise-disjoint, bounded, connected, simply connected, open subsets of $\mathbb{R}^2$. Let $A = U \cup V$ and $B = Y \cup Z$ be bounded open subsets each with two components. Suppose we also know the following, where '$\simeq$' refers to planar isotopy:
$U \simeq Y$ (which extends to a $\partial (U) \simeq \partial (Y)$)
$V \simeq Z$ (which extends to a $\partial (V) \simeq \partial (Z)$)
$\bar{U} \cap \bar{V} \simeq \bar{Y} \cap \bar{Z}$
Then this is not enough to imply that $A \simeq B$, due to the following example. Let $U = Y$ be a 'thickened open topologist's sine curve', i.e. a tubular neighborhood of the standard topologist's sine curve with limit arc being $[-1, 1]$ on the $y$-axis (though not itself containing this arc, just the 'wiggly part' on $(0, \pi)$). Let $V$ be the reflection of $U$ across the $y$-axis, and let $Z$ be the same as $V$ except scaled by $\frac{1}{2}$ vertically. So in other words $Z$ is just a vertically-squished topologist's sine curve coming from the other direction. Then clearly $U, V, Y, Z$ satisfy all of the above conditions (condition 3 is satisfied since each is just an arc, albeit of different lengths) but $A \not\simeq B$.
So I'm wondering, what are some of the simplest conditions we can add on to those three above to ensure that $A \simeq B$? Is it enough that $\bar{A}$ is homeomorphic to $\bar{B}$? It feels like this is probably enough at least in our simplified case, but I'd be wary of assuming it's enough after removing the simply-connected condition, and obviously in the unbounded case as well. Perhaps, is there something weaker than just directly assuming homeomorphism that still gets you there in our case?
Another difference in our case is that $\bar{U} \setminus \bar{V}$ is not homeomorphic to $\bar{Y} \setminus \bar{Z}$, since in one case the limit arc is completely gone, while in the other case two pieces of it remain. You can use a basic local connectedness argument at one of these remaining 'bar' points to show that there is no homeomorphism in case it's not clear. Would assuming this, as well as its obverse for $\bar{V} \setminus \bar{U}$ and $\bar{Z} \setminus \bar{Y}$, be sufficient? I'm more skeptical of this, but who knows.
Then, obviously there are a few generalizations you could make by removing either or both of 'bounded' or 'simply connected,' or you could allow $A$ and $B$ to have $n > 2$ components and having some combinatorial collection of $k$-wise component/boundary isotopies as $k$ runs from $2$ to $n$ (though this might derive from the $n=2$ case directly). You could also try to generalize to infinitely many components with $A$ and $B$ still bounded, but that is probably a whole different ball of wax when talking about isotopy.
Is anyone aware of the state of the art, or perhaps even a solution, to this sort of problem?