Define an inclusion of a polygon $A$ to be a subset of the complement of the polygon whose boundary minus $A$ consists of a single edge. Inclusions sometimes contain other inclusions. The congruence class of an inclusion is the congruence class of the pair (polygon, free edge).
Define the inclusion vector $Iv(A)$ to be the nonngegative linear combination of congruence classes of inclusions of $A$ describing all inclusions up to congruence.
Notice that for any $X$ in $A + P$, as long as neither $A$ nor $P$ fit inside an inclusion of the other, then $Iv(X) \ge Iv(A) + Iv(B)$
Let's say that a polygons $X$ and $Y$ bind to each other with strength $k$ if they can be arranged up to congruence so that their intersection is a connected set of $k > 2$ edges. Then any inclusions $J$ of an element $X$ of $A + P$ that are not inclusions of $A$ or of $P$ must be polygons that bind to $A$ and $P$ with total strength at least the (length of the perimeter of $J$) - 1.
With the picture established by all these definitions, it's easy to make counterexamples. For $A$ and $B$, start with large rectangles with crenelated boundaries (every other square along the boundary removed). There are necessarily 4 straight segments of length 2 on the boundary at the 4 corners, but that is all. Now bore out at least 5 maximal inclusions from $A$ and $B$ in the form of $S$-patterned double spirals that do not bind to $A$ or $B$ anywhere except at the four straight stretches. Make sure the maximal inclusions removed from $A$ and $B$ are not congruent, and that the resulting polygons $A'$ and $B'$ have the same area.
The difference between the vectors $Iv(A)$ and $Iv(B)$ is sufficiently large that adding a polygon $P$ can never create enough inclusions to bridge the difference.