Let $AC$ intersect $BX$ at $P$, and let $AB$ intersect $CY$ at $Q$, as shown. From $\triangle XBA\cong \triangle ACY$, the fact that $\angle XBA=\angle ACY$ implies that $CBPQ$ is cyclic. Now let $O$ be the point on this circle such that $OB=OC$. We will now show that $O$ is the circumcenter of $\triangle XAY$. If this is true, then since $O$ lies on the perpendicular bisector of $BC$, it would follow that $OA'=OA$ and thus $A,A',X,Y$ would lie on a circle with center $O$. Firstly, the length conditions and the equality $\angle OBX=\angle OCA$ (from $CBPO$ cyclic) imply that $\triangle OXB\cong \triangle OAC$, so $OX=OA$. Also, the length conditions and the equality $\angle OBA=\angle OCY$ (from $CBQO$ cyclic) imply that $\triangle OBA \cong \triangle OCY$, so $OA=OY$. Since $OX=OA=OY$, $O$ is indeed the circumcenter of $\triangle XAY$, and we are done.

Solution. WLOG, $AC\geq AB$. Let $O$ be the intersection point of the perpendicular bisectors of segments $BC$ and $AX$. Clearly, $OA = OA'$. Since $OB= OC,\ OX = OA$ and $XB = AC$ we have $\bigtriangleup BOX \cong \bigtriangleup COA$. Hence $$\angle OCY = \angle OCA - \angle YCA = \angle OBX -\angle ABX = \angle OBA$$which together with $CY = BA$ and $OB = OC$ implies that $\bigtriangleup ABO \cong \triangle YCO$, i.e. $OA = OY$. Summarizing, we have obtained that $OA' = OA = OX = OY$, which leads to the required result. $\square$

Center of rotation $ABX\mapsto YCA$ lay on perp bisector of $BC$ (since $B\mapsto C$). Result is obvious now. Edit: Someone asked me to explain more the last sentence, so I better do it here. Center of rotation is center of $(XAY)$ and by symmetry $A'$ also lays on this circle.

If $Y'$ is reflection of $Y$ with respect to perpendicular bisector of $BC$, then $\triangle BY'A' \equiv \triangle BAX$ so $XAY'A'$ is isosceles trapezoid. $AY'YA'$ is also isosceles trapezoid, therefore because $XAY'A'$ and $AY'YA'$ are cyclic we know that $XAYA'$ is cyclic too.