Let $Z$ be the midpoint of the longer arc $BC$ of the circumcircle $\omega$ of the triangle $ABC$. The triangles $ZDB$ and $ZEC$ are congruent, because they agree in the sides $BD = CE$ and $ZB = ZC$, as well as in the corresponding angles between them, for both lie over the chord $AZ$ of $\omega$. It follows that $\angle ZDA = \angle ZEA$, which in turn discloses that the quadrilateral $ADEZ$ is cyclic. So it remains to be shown that the quadrilateral $AO_1O_2Z$ is cyclic.

The center $O$ of $\omega$ satisfies $OO_1 \perp AB$ and $OO_2 \perp AC$. The projections of $O$ and $O_1$ onto $AC$ are the midpoints of $AC$ and $AE$ respectively. Thus the projection of the segment $OO_1$ onto $AC$ has length $\frac{1}{2}CE$. For the same reason, the projection of $OO_2$ on $AB$ has length $\frac{1}{2}BD$, and by hypothesis these two length agree. Moreover, the angle between $OO_1$ and $AC$ is the same as the angle between $OO_2$ and $AB$. It follows that $OO_1 = OO_2$.
Further, we have $\angle AOO_1 = \angle ACB = \angle O_2OZ$, the latter being a consequence of $ZO \perp BC$ and $OO_2 \perp AC$. So the rays $OA$ and $OZ$ are isogonal in the angle $O_2OO_1$. In the combination with $AO = ZO$ and $OO_1 = OO_2$ this proves that

the quadrilateral $AO_1O_2Z$ is an isosceles trapezium and thus in particular cyclic.
Thereby the problem is solved.

Let the circumcircles of triangles $ABE$ and $ADC$ intersect each other again at $F \neq A$. Then the triangles $BFD$ and $EFC$ are congruent, for they agree in their sides $BD = CE$ as well as in their corresponding adjacent angles, i.e., $\angle FBD = \angle FEC$ and $\angle BDF = \angle ECF$. It follows that the altitudes of these triangles passing through $F$ have the same lengths, wherefore $AF$ is the bisector of the angle $BAC$.

Now construct the point $S$ such that $SO_1FO_2$ is a parallelogram. We will show that $S$ is the desired point.
To prove that $S$ lies on the circumcircle of triangle $AO_1O_2$, we note that the triangles $AO_1O_2$ and $FO_1O_2$ are congruent due to $AO_1 = FO_1$ and $AO_2 = FO_2$. It follows that $AO_1O_2S$ is an isosceles trapezium and hence in particular a cyclic quadrilateral, as claimed. Later, it will help us to have observed that the facts used in this paragraph imply $AS \parallel O_1O_2 \perp AF$.
Next, we prove that $S$ lies on the circumcircle of the triangle $ABC$ and that it is actually the midpoint of its longer arc $BC$; this will also show $S \neq A$, as needed. Our first intermediate step is to observe that the triangles $O_1SB$ and $O_2CS$ are congruent. Indeed they agree in a pair of sides, $BO_1 = FO_1 = SO_2$ and $SO_1 = FO_2 = CO_2$. Moreover the corresponding angles between these sides are equal, because their complements to $360^\circ$ are equal as a consequence of
$$\angle BO_1F = 2\angle BAF = 2\angle FAC = \angle FO_2C$$
and $\angle FO_1S = \angle SO_2F$. This concludes the verification of $\triangle O_1SB \cong \triangle O_2CS$, and it follows that $BS = CS$. Further, since $AF$ is the bisector of $\angle BAC$ and $AF \perp AS$, the line $AS$ is the exterior bisector of $\angle BAC$. Altogether we obtain that $S$ is the point described above. The fact that $S$ lies on the circumcircle of the triangle $ADE$ can be shown as in the first solution.

Denote by $O$ and $P$ the circumcentres of triangles $ABC$ and $ADE$, respectively. The lines $OO_1$ and $O_2P$ (being the perpendicular bisectors of $AB$ and $AD$, respectively) are both perpendicular to $AB$ and their distance is $\frac{1}{2}BD$. Similarly, the lines $OO_2$ and $O_1P$ are both perpendicular to $AC$ and their distance is $\frac{1}{2}CE$. Since $\frac{1}{2}BD = \frac{1}{2}CE$, the quadrilateral $O_1OO_2P$ is a parallelogram with equal altitudes, hence a rhombus. It follows that $OP$ is the perpendicular bisector of $O_1O_2$, so all the three circumcentres of the triangles $ABC$, $ADE$ and $AO_1O_2$ lie on the same line, which concludes the claim (since $A$ does not lie on this line because of $AB \neq AC$).