A polyomino puzzle

${}{}{}{}{}{}{}{}{}{}{}{}{}{}$

This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:

Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n \in \mathbb{Z}$. (Yes $\mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )

A similar construction works for any pair of finite polyominos.

By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).

If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.

$\quad\square$
$\square\square\square$
$\quad\square$
$\quad\square$
$\quad\square$