Claim 1. $AE$ and $AF$ are symmetric with respect to $AI$.
Proof. Let $O$ be the circumcenter of $ABC$. By definition of $F$, $OE \cdot OF = OA^2$. Hence $\triangle OEA \sim \triangle OAF$. It follows that $\angle EAO = \angle OFA$. Hence
$$\angle FAM = \angle OMA - \angle MFA = \angle MAO - \angle EAO = \angle MAE.$$
Claim 2. $\dfrac{BG}{CG}=\dfrac{BA}{CA}$.
Proof. By Claim 1., $\angle BAG = \angle EAC$. Also, $\angle AGB = \angle ACE$. Hence $\triangle ABG \sim \triangle AEC$. Similarly $\triangle GAC \sim \triangle BAE$. It follows that
$$\frac{AB}{BG} = \frac{AE}{EC} = \frac{AE}{EB} = \frac{AC}{CG},$$
so $\dfrac{BG}{CG}=\dfrac{BA}{CA}$.
Claim 3. Let $BI$, $CI$ intersect the circumcircle of $ABC$ at $X$, $Y$, respectively. Then
$$\frac{XD}{YD}=\frac{XM}{YM}.$$
Proof. Since $\triangle BIG \sim \triangle DIX$, we have
$$\frac{BG}{BI} = \frac{XD}{DI}.$$
Analogously,
$$\frac{CG}{CI} = \frac{YD}{DI}.$$
Hence
$$\frac{XD}{YD}=\frac{BG}{CG} \cdot \frac{CI}{BI}.$$
Analogously
$$\frac{XM}{YM}=\frac{BA}{CA} \cdot \frac{CI}{BI}.$$
But by Claim 2. we have $\dfrac{BG}{CG}=\dfrac{BA}{CA}$, so
$$\frac{XD}{YD}=\frac{BG}{CG} \cdot \frac{CI}{BI}=\frac{BA}{CA} \cdot \frac{CI}{BI} = \frac{XM}{YM}.$$
Claim 4. $MX$ and $MY$ bisect the segments $CI$, $BI$, respectively.
Proof. By Trefoil Lemma, $XI=CI$ and $MI=CI$. Hence $MX$ is perpendicular bisector of $CI$, hence it bisects $CI$. Analogously, $MY$ bisects $BI$.
Claim 5. Let $K$, $L$ be the midpoints of $CI$, $BI$. Let $DM$ instersect $KL$ at $Z$. Then $Z$ is the midpoint of $KL$.
Proof. Let $\ell$ be the tangent to the circumcircle of $ABC$ at $M$. Then $\angle MYD = \angle(\ell, MD) = \angle(KL, MD) = \angle LZM$. Hence $\triangle MYD \sim \triangle MZL$. Analogously, $\triangle MXD \sim \triangle MZK$. It follows that
$$\frac{MZ}{ZL} = \frac{MY}{YD} = \frac{MX}{XD} = \frac{MZ}{ZK},$$
so $ZL=ZK$.
Claim 6. $Z$ is the midpoint of $IE$.
Proof. $KILE$ is a parallelogram since $E,K,L$ are the midpoints of $BC$, $CI$, $IB$. Hence midpoints of $KL$, $IE$ coincide. Hence the claim.
The problem is solved by Claim 6.
The following is a solution using projective geometry.
Let $X$ and $Y$ be the midpoints of arcs $AC$, $AB$ of the circumcircle of $ABC$. Then $X$ lies of $BI$ and $Y$ lies on $CI$. Moreover, due to well-known trefoil lemma, $MB=MI=MC$, $XC=XI$ and $YB=YI$. Hence $XM$ and $YM$ intersect $CI$ and $BI$ at their midpoints $K$ and $L$, respectively.
Using well-known properties of symmedians we see that $(AGBC)=-1$.
It is well-known that given a circle $\omega$ and a point $P$, the map $\phi \colon \omega \to \omega$ assigning to a point $T$ the unique point $T'\in \omega$ such that $T,P,T'$ are collinear is projective. Taking as $\omega$ the circumcircle of $ABC$ and $I$ as a $P$ we get that
$$-1=(AGBC)=(\phi(A)\phi(G)\phi(B)\phi(C))=(MDXY).$$
Projecting $\omega$ through $M$ to the line $KL$ we obtain
$$-1=(MDXY)=(\infty ZKL).$$ where $\infty$ is the point at infinity of the line $KL$ and $Z$ is the intersection of $KL$ with $DM$. This means that $Z$ is the midpoint of $KL$.
But $E,K,L$ are the midpoints of $IBC$, so $IKEL$ is a parallelogram, hence $Z$ is a midpoint of $IE$. We are done!
What about this solution (instead of searching for a credible and/or official solution that you may never find to give it the bonus reputation)?
Like the capitals you've associated to points, let $I$ be the center of inscribed circle inside $\bigtriangleup ABC$. Take $S$ the intersection of $AD$ and $BC$, then we have:
$$\dfrac {AI} {IS} = \dfrac {\sin \angle DAB}{\sin \angle CSA}= \dfrac {\sin \angle SCD}{\sin \angle CSD}= \dfrac{DC}{DS} =\dfrac {DI} {DS}$$
Let $P$ be the intersection of line $\overline {IE}$ and $DL$ and $Q$ be the intersection of $IH$ and $DF$. By Seva's theorem in $\bigtriangleup IDH$ and $IP \| AH $ and above equation we have:
$$\dfrac {IQ}{QH}= \dfrac {PD}{HP} . \dfrac {IS}{DS}= \dfrac {DI}{AI} . \dfrac {AI}{DI}=1$$
So $Q$ is the midpoint of $IH$. let $R$ the intersection of $AH$ and $DF$, then $IEHR$ is a rectangle. From the fact that the quadrilateral $ASEF$ is cyclic (why?) and $RI \| BC$, we conclude $AIRF$ is cyclic, so as you mentioned $IF \bot AK$, therefore $IEKF$ and then $AIHK$ is cyclic and we have $KI\bot AD$ and $\angle HID = \angle AKJ$.
Ultimately, $\bigtriangleup HID \sim \bigtriangleup JKF$ and the tangent of $(ABC)$ at $F$ bisects $KJ$ like the way $DF$ bisects $IH$ at $Q$.
I think other solutions may use the cyclic characteristics of quadrilaterals $ASEF$ and $FEHL$.
Best Answer
Your attempt is on the right track, though somewhat in reverse.
For the sake of completeness, I will post the full solution.
It is well known by the incenter-excenter lemma (see a proof here) that $BI_ACI$ is cyclic with diameter $II_A$. Using the notation in OP, here is another lemma:
It is not hard to see the finish from here: