Let $O_C$ be the circumcenter of triangle $AGB$ and $O$ be the midpoint of $GO_C$. We claim that $O$ is the desired common center. Let $X_1$ and $Y_1$ be the midpoints of $AX$ and $BY$, respectively. To see why, first note that $OA=OD$ and $OB=OE$ follows by dilation at $G$. Now, we see that $GX=GY=GF$ and $\angle XGY=2\angle AGB$; so in fact, we have $\triangle XGY \sim \triangle AO_CB$. Let $T=AX \cap BY$; then it is clear that $T \in (XGY)$ and $(AO_CB)$. Finally, consider the spiral similarity $\mathcal{S}: AB \mapsto XY$; viewing it as a dynamic process; by the Averaging Principle, we see that $$\triangle X_1OY_1 \sim \triangle AO_CB \sim \triangle XGY.$$Consequently, we have $T \in (X_1OY_1)$ and $OX_1=OY_1$. As $\angle OX_1A=\angle OY_1Y$ and $AX_1=YY_1=\tfrac{1}{4}AB$, we conclude that $\triangle OX_1A \sim \triangle OY_1Y$. Hence, we get $\triangle OXA \sim \triangle OBY$. As $AX=BY=\tfrac{1}{2}AB$, the similarity is a congruency, so $OA=OY$ and $OX=OB$ as desired. $\blacksquare$

Since $BE \parallel YD$ and $BY=DE$, we know that quadrilateral $BYDE$ is an isosceles trapezoid and is consequently cyclic; it follows that the line between $BE\cap YD$ and $AD\cap XE$ is the radical axis of $\odot(BEX)$ and $\odot(ADY)$, which is clearly the line at infinity, so the two circles must be concentric.

diagram

Attachments:

Obviously $BYDE$ and $AXED$ cyclic. But since, $AX=ED$ and $BY=DE$ $\implies$ $BYDE$ and $AXDE$ are isosceles trapezium $\implies$ perpendicular bisectors of $XE,AD$ and $YD,BE$ are same

Denote by $\measuredangle$ as the angle $\angle$ modulo $\pi$. Make the following observations that $XA = ED$ and $BY = DE$. Indeed, \[XA = AF = FB = ED, \]and a similar process for $BY = DE$. Claim: $XEDA$ and $DYBE$ are isosceles trapezoids. Proof: We prove the case for $XEDA$, as the $DYBE$ case is a nearly identical. Observe that \[\measuredangle DAX = \measuredangle FAD = \measuredangle EDA,\]and as $XA = ED$, we must have $XEDA$ is an isosceles trapezoid (as long as $X \neq E$). $\square$ Now observe that the perpendicular bisectors of $XE, AD$ and $YD, BE$ are identical as $XE \parallel AD$ and $YD \parallel BE$. Now the point at which these perpendicular bisectors intersect will become our desired circumcenter of $BEX$ and $ADY$, so that their circumcircles are concentric, as desired. $\blacksquare$

Let $M$ be the midpoint of $AD$ and $N$ the midpoint of $BE$. Define $O$ such that $OM$ is perpendicular to $AD$ and $ON$ is perpendicular to $BE$ (notably, $AO = OD, BO = OE$). We claim that $DO = OY$ (similarly $EO = OX$); this would solve the problem. Indeed, note that $NF = NY = ND$, and $ON, FY$ are both perpendicular to $NE, YD$. So $ON$ is the perpendicular bisector of $YD$ as desired.