Here's an incomplete proof. An equivalent question is whether a polygon $P$ with $180^\circ$ rotational symmetry about $O$ can be divided into two congruent polygons such that the dividing curve $D$ between the polygons does not have $180^\circ$ symmetry about $O$.
First, we can rule out $D$ having $180^\circ$ symmetry about a point $O'\neq O$. This is because, in such a case, we can shift $D$ to have it's center at $O$ (perhaps extending $D$ in a symmetric way if necessary), which will result in a new division of $P$ into $A'$ and $B'$. By the $180^\circ$ symmetry of $P$ and the new shifted $D$ about $O$, $A'$ and $B'$ must have the same area. However, this shifting must've increased the area of one of $A,B$ (and correspondingly decreased the other), meaning $A,B$ had different areas. But this contradicts them being equivalent.
So now suppose we have a $D$ that is not symmetric about any point. $D$ then looks different whether you are approaching $D$ from the $A$ side or the $B$ side. Let's call the two versions "innie" and "outie" respectively.
Now, each of $A$, $B$, $P$ may have several copies of $D$ along their respective boundaries, either as innies or outies. Suppose for the moment that the various innies and outies cannot overlap except possibly at their endpoints. Then the number of innies of $P$ and outies of $P$ must each be finite, and moreover must each be even, by rotational symmetry. The total number of innies between $A$ and $B$ is therefore odd, and the total number of outies odd as well, since the boundary $D$ contributes one more innie to $A$ and outie to $B$ that were not included in the boundary of $P$. But this is impossible: since $A$ and $B$ are equivalent, they have the same number of innies (and same number of outies) and therefore the total number of innies and outies between them must each be even.
It remains to handle the case where the copies of the dividing curve $D$ found along the boundaries can intersect each other. There are a couple potential issues. The first is that the number of copies along the boundaries may be infinite. This can only occur, however, if $D$ is a line segment (in which case a longer line segment somewhere on the boundary would contain infinitely many copies of $D$). But line segments have rotational symmetry, which we've already ruled out.
The bigger problem is that you may have a copy $D$' of $D$ along the boundary of $A$ or $B$, such that part of $D'$ overlaps $D$, and part does not. Then $D'$ is counted in the count of innies or outies of $A$, but not in the count of innies or outies of $P$. I suspect you can handle this by breaking $D$,$D'$ into the segments of overlap and non-overlap and doing a similar counting argument, but I haven't found a concise proof of this yet.