Let $F$ be the point such that $AECF$ is a parallelogram, and let $G$ be the point such that $AGBE$ is a parallelogram. Then $FGBC$ is also a parallelogram. Then $\measuredangle AFG = \measuredangle ECB$ (since $FG$ parallel to $BC$, $AF$ parallel to $EC$) $= \measuredangle BAE = \measuredangle ABG$ so $AGBF$ is a cyclic quadrilateral. So $\measuredangle EBA = \measuredangle GAB = \measuredangle GFB = \measuredangle CBF$. Then since $BE$ is shared and $\measuredangle BAE = \measuredangle ECB$, circles $ABE$ and $ECB$ have the same radius. Also, $AE$ is shared, and $\measuredangle ADE = \measuredangle CBF = \measuredangle EBA$ so circles $ABE$ and $ADE$ have the same radius. Similarly, circles $CDE$ and $ECB$ have the same radius. Therefore, circles $ABE$, $ECB$, $CDE$, $DAE$ have the same radii, and their centres therefore lie on a circle with this same radius, centre E.

Let the perpendicular from $E$ to $CD,AB,AD,BC$ intersect them at $K,L,M,N$ respectively. $\angle LKN=\angle EKN=\angle ECN=\angle LAE=\angle LME=\angle LMN$ Thus $KLMN$ is cyclic. $\angle CBE=\angle NBE=\angle NLE=\angle NLK=\angle NMK=\angle EDK=\angle EDC$ $\frac{BE}{\sin ECB}=\frac{BE}{\sin BAE}\implies EO_1=EO_2$ $\frac{ED}{\sin DCE}=\frac{ED}{\sin EAD}\implies EO_3=EO_4$ $\frac{EC}{\sin EDC}=\frac{EC}{\sin CBE}\implies EO_3=EO_2$ Thus $EO_1=EO_2=EO_3=EO_4$ which gives that $O_1,O_2,O_3,O_4$ lie on a circle whose circumcenter is $E$ as desired.$\blacksquare$