Let $M$ be the midpoint of the arc $AB$ of $\omega_1$ and redefine $H$ as the projection of $M$ on $CF.$ It suffices to show that $M,H,G$ are collinear. Label $\alpha,\beta,\gamma$ the angles of $\triangle ABC$ at $A,B,C.$ $\angle CMH=90^{\circ}-(\angle MCE-\angle FCE)=90^{\circ}-\tfrac{1}{2}\beta-\tfrac{1}{2}\gamma-\angle FCE$ $\Longrightarrow$ $\angle CMH=\frac{_1}{^2}\alpha-\angle FCE.$ Let $P \equiv CD \cap EF.$ From cyclic $CDEF,$ we have $\angle PFC=\angle FEC+\angle FCE=\angle CDA=\frac{_1}{^2}\alpha,$ but since $\angle GFC=\angle FEC$ $\Longrightarrow$ $\angle GFC=\frac{_1}{^2}\alpha-\angle FCE=\angle CMH$ $\Longrightarrow$ $G \in MH,$ as desired.

This is straightforward angle chasing, so I won't post my solution. Here are some remarks. Let $O_2$ be the center of $\omega_2$. Since the perpendicular bisectors of $AD$ and $AE$ meet at $O_2$ and are the external angle bisectors of $\angle B$ and $\angle C$, it follows that $O_2$ is the $A$-excenter of triangle $ABC$. In particular, line $OO_2$ is the perpendicular bisector of segment $AF$, so $F$ can be reinterpreted as the reflection of $A$ over line $I_aO$. In fact, I think this makes a harder restatement of the problem.

Let $O$ be the circumcentre of $\triangle ABC$ and $M$ be the midpoint of arc $AB$ not containing $C$. Note that since $\triangle DAC$ and $\triangle EBC$ are isosceles, the external angle bisectors of $\angle A, \angle B$ are the perpendicular bisectors of $DC, EC$, respectively. Hence the circumcentre of $\triangle CDE$ is $I_c$, the excentre of $\triangle ABC$ opposite $C$. Now it is easy to complex with unit circle $(ABC)$ and $a^2,b^2,c^2$ being the vertices $A,B,C$. $m=-ab, i_c=-ab+bc+ca$. $\ [ \begin{align*}CF \perp I_c O \implies -cf &= -\frac{i_c}{\overline{i_c}}\\ &= -\frac{abc(-ab+bc+ca)}{a+b-c} \end{align*}$ $\implies f=\frac{ab(-ab+bc+ca)}{c(a+b-c}.$ So $f - i_c = (-ab+bc+ca)\frac{ab-c(a+b-c)}{c(a+b-c)} = \frac{(-ab+bc+ca)(c-a)(c-b)}{c(a+b-c)}. $ $\overline{(f-i_c)} = \frac{(a+b-c)(c-a)(c-b)}{(-ab+bc+ca)abc}$. So $\frac{f-i_c}{\overline{(f-i_c)}} = \frac{(-ab+bc+ca)^2ab}{(a+b-c)^2}$. This last quantity equals $fg$ since $I_cF \perp FG$ so $g=\frac{c(-ab+bc+ca)}{a+b-c}$ from which we observe that $mg=-c^2f$. So $MG \perp CF$ and we are done.

Let $I$ be the incenter of $\bigtriangleup ABC$, $M$ be the midpoint of arc $AB$ of $\omega_1$, and $K=AB\cap MC$. Since $AI\parallel DC$, $BI\parallel EC$, $\bigtriangleup AIB, \bigtriangleup DCE$ are homothetic with center $K$. It is well-known that the circumcenters of $\bigtriangleup AIB, \bigtriangleup DCE$ are $M,O$ respectively, so this homothety also sends $M$ to $O$, thus $K,M,O$ are collinear. But $C,K,M$ are collinear, thus we actually have $C,M,O$ collinear. Now we have \[\angle FCM+\angle GMC=\angle OFC+\angle CFG=90^\circ \Rightarrow CF\perp GM\] Thus $G,H,M$ are collinear. But $GM$ is the bisector of $\angle AGB$, so in fact $GH$ bisects $\angle AGB$, and thus $\angle AGH=\angle BGH$.

First note that $\angle AFD=\angle AFB - \angle DFC=\frac{\angle B}{2}$. Let $\angle GFC=\phi$. From that we have $\angle CDF=\phi$ so $ \angle ADF= \frac{\angle A}{2} - \phi$. From this we easily get $\angle FCB=\angle FAB= 90- \frac{\angle C}{2} - \phi $. Now let $M$ be the midpoint of arc $AB$ containing $C$ we have that $\angle BAM=90-\frac{\angle C}{2}$ which implies $\angle FCM=\phi=\angle GFC$ so $CFGM$ is isosceles trapezoid so $MG || CF$ so $GH \perp MG$ so $\angle AGH=\angle BGH$ so we are finished.

If D,E on AB the problem is true

[asy][asy]size(10cm); pair A=(0,0),B=(10,0),C=(-1,5),D,Ee,F,G,H,M,Nn; D=abs(C-A)*W;Ee=B+abs(C-B)*E; F=intersectionpoints(circumcircle(A,B,C),circumcircle(C,D,Ee))[0]; M=extension(C,bisectorpoint(B,C,A),(B+A)/2,bisectorpoint(B,A)); Nn=2*circumcenter(A,B,C)-M; G=reflect((C+F)/2,bisectorpoint(C,F))*Nn;H=foot(G,C,F); draw(arc(circumcenter(C,D,Ee),Ee,D)^^circumcircle(A,B,C),green); D(A--G--B,heavycyan+dashed); D(MP("A",A,S)--MP("C",C,NW)--MP("D",D,S)--MP("B",B,S)--MP("E",Ee,S)--C--B); D(C--MP("F",F,N)--MP("G",G,NW)--MP("H",H,SE)--MP("M",M,S)--C--MP("N",Nn,N),magenta); dot(A^^B^^C^^D^^Ee^^F^^M^^Nn^^G^^H); [/asy][/asy] Let $GH$ meet $\omega _1$ again at $M$ and let the tangent to $\omega_2$ at $C$ meet $\omega_1$ again at $N$. Note that $$\angle NCB=\angle NCE+\angle ECB=\angle CDA+\angle ECB=\tfrac12\angle CAB+\tfrac12\angle CBA=90-\tfrac12\angle ACB.$$So $CN$ is the external bisector of $\angle ACB\implies N$ is the midpoint of arc $\widehat{ACB}$. Also, $G,N$ are symmetric w.r.t. the perpendicular bisector of $CF$, so $\angle NCF=\angle GFC$. Therefore, $$\angle NCM=\angle NCF+\angle FCM=\angle GFC+\angle FGH=90^\circ.$$So $M$ is the antipode of $N$, i.e., the midpoint of arc $\widehat{AB}$. So $$\angle AGH=\angle AGM=\angle BGM=\angle BGH,$$as required. $\blacksquare$