We claim that $O_{AB}O_{DE}$, etc, pass through the midpoint of $OX$. Note that $\triangle{BOD} \sim \triangle{BO_{AB}X} \sim \triangle{XO_{DE}D} \implies \triangle{DXB} \sim \triangle{DO_{DE}O} \implies OO_{DE} = DO_{DE} \cdot \frac{BX}{DX} = XO_{AB} \implies O_{AB}XO_{DE}O$ is a parallelogram, done. $\square$

Very nice problem! Here's a solution using inversion. Let the reflections of $P$ over lines $O_{AB}O_{DE}, O_{BC}O_{EF}, O_{CD}O_{FA}$ be $P_1,P_2,P_3$ respectively. Simple angle chasing yields $O,P$ are isogonal conjugates in hexagon $O_{AB}O_{BC}O_{CD}O_{DE}O_{EF}O_{FA}$, and so $O_{AB} O_{DE}, O_{BC}O_{EF}, O_{CD}O_{FA}$ are concurrent (at a point, say $T$) by Brianchon's theorem. Note that $T$ is the center of $\odot (PP_1P_2P_3)$ since $TP=TP_1=TP_2=TP_3$. Claim: The radical axis of $\odot(PP_1P_2P_3)$ and $\odot(O)$ coincides with the polar of $X$ wrt. $\odot (O)$ (denote by $l$). Proof: We only need to prove that there exists a circle $\omega$ through $A, D$ orthogonal to both $\odot(PP_1P_2P_3)$ and $\odot(O)$. (Then we are done since $AA\cap DD, BB\cap EE, CC\cap FF$ all lie on $l$.) Now invert about $P$. Since $(A,D), (B,E), (C,F)$ are pairs of an involution on $\odot (O)$, the hexagon $A'B'C'D'E'F'$ obtained after inversion also satisfies the condition $A'D', B'E', C'F'$ are concurrent. From here we may use notations from the original problem. It suffices to show that the circle $\omega$ through $A,D$ orthogonal to $\odot (O)$ is also orthogonal to $l$, but the center of $\omega$ is exactly $AA\cap DD$ which lies on $l$. $\blacksquare$ From the claim we get $OX \perp l$, $OT \perp l$, hence the final line $OX$ is also through $T$. $\blacksquare$