Solution:

If $X$ is the center of the circumcircle of $\triangle ODE$, then $90^\circ - \angle KDE = 90^\circ - \angle ODE = \angle XEO = \angle XEL = \angle XHD = \angle XKD$ (all angles are oriented), from which it follows that $XK \perp DE$; analogously $XL \perp DE$, i.e. $K$ and $L$ lie on the perpendicular bisector of segment $DE$, so $DEHO$ is an isosceles trapezoid, hence $D, E, O, H$ lie on a circle.

On the other hand, if $O$ lies on the circle $HDE$, that is, on the circle with diameter $CH$, the inscribed angles over $EH$ and $OD$ are equal ($\angle ECH = \angle OCD$), so $DEHO$ is an isosceles trapezoid, and from this $DL = EL$. We now have $\angle EXH = \angle ELH = 2 \angle EDH$

and analogously $\angle DXH = 2 \angle DEH$, from which it follows that $X$ is the center of the circle $DEOH$. Therefore, $X$ is the center of the circle $ODE$ if and only if $D, E, O$ and $H$ lie on a circle.

If the points $D, E, O, H$ lie on a circle, then $K, L$ and $M$ lie on the perpendicular bisector of segment $DE$, which proves one direction of the problem. Now suppose that $O$ lies outside the circle $CDHE$ (the case when $O$ is inside the circle is treated in the same way). Since $CO \perp DE$, we have $DL > LE$ and $EK > KD$, i.e. $K$ and $L$ lie on different sides of the perpendicular bisector of segment $DE$, while $M$ belongs to this perpendicular bisector. Therefore, if $K, L$ and $M$ are collinear, $M$ must lie between $K$ and $L$. It follows that one of the points $K$ and $L$ is outside triangle $ABC$, and the other is inside the triangle. However, when $O$ is outside the quadrilateral $ABDE$, both points $K$ and $L$ are outside the triangle, and otherwise both are inside the triangle. This is a contradiction with the assumption that $M$ lies on the line $KL$, which proves the other direction.