If $AA_1 = CC_1$, then the hexagon is symmetric about the line $BB_1$; in particular the circles $ABC$ and $A_1BC_1$ are tangent to each other. So $AA_1$ and $CC_1$ must be different. Since the points $A$ and $A_1$ can be interchanged with $C$ and $C_1$, respectively, we may assume $AA_1 < CC_1$.
Let $R$ be the radical center of the circles $AEBC$ and $A_1EBC_1$, and the circumcircle of the symmetric trapezoid $ACC_1A_1$; that is the common point of the pairwise radical axes $AC, A_1C_1$, and $BE$. By the symmetry of $AC$ and $A_1C_1$, the point $R$ lies on the common perpendicular bisector of $AA_1$ and $CC_1$, which is the external bisector of $\angle ADC$.
Let $F$ be the second intersection of the line $DR$ and the circle $ACD$. From the power of $R$ with respect to the circles $\omega$ and $ACFD$ we have
$$RB \cdot RE = RA \cdot RC = RD \cdot DF,$$
so the points $B, E, D$ and $F$ are concyclic.
The line $RDF$ is the external bisector of $\angle ADC$, so the point $F$ bisects the arc $CDA$. By $AB = BC$, on circle $\omega$, the point $B$ is the midpoint of arc $AEC$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite arc $CA$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $AC$, so they are collinear.

Finally, let $X$ be the second intersection point of $\omega$ and the line $DE$.
Since $BM$ is a diameter in $\omega$, we have $\angle BXM = 90^\circ$. Moreover,
$$\angle EXM = 180^\circ - \angle MBE = 180^\circ - \angle FBE = \angle EDF,$$
so $MX$ and $FD$ are parallel. Since $BX$ is perpendicular to $MX$ and $BB_1$ is perpendicular to $FD$, this shows that $X$ lies on the line $BB_1$.