Solution:
Since $CD = CE$ it means that $E$ is the reflection of $D$ on the bisector of $\angle ACB$, i.e. the line $CI$. Let $G$ be the reflection of $F$ on $CI$. Then $G$ lies on the segment $CE$, the segment $EG$ is the reflection of the segment $DF$ on the line $CI$. Also, the quadrilateral $DEGF$ is cyclic since $\angle DFE = \angle EGD$.

Suppose that the quadrilateral $ABEF$ is circumscribable. Since $\angle FAI = \angle BAI$ and $\angle EBI = \angle ABI$, then $I$ is the centre of its inscribed circle. Then $\angle DFI = \angle EFI$ and since segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle EFI = \angle DGI$. So $\angle DFI = \angle DGI$ which means that quadrilateral $DIGF$ is cyclic. Since the quadrilateral $DEGF$ is also cyclic, we have that the quadrilateral $DIEF$ is cyclic.

Suppose that the quadrilateral $DIEF$ is cyclic. Since quadrilateral $DEGF$ is also cyclic, we have that the pentagon $DIEGF$ is cyclic. So $\angle IEB = 180^{\circ} - \angle IEG = \angle IDG$ and since segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle IDG = \angle IEF$. Hence $\angle IEB = \angle IEF$, which means that $EI$ is the angle bisector of $\angle BEF$. Since $\angle IFA = \angle IFD = \angle IGD$ and since the segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle IGD = \angle IFE$, hence $\angle IFA = \angle IFE$, which means that $FI$ is the angle bisector of $\angle EFA$. We also know that $AI$ and $BI$ are the angle bisectors of $\angle FAB$ and $\angle ABE$. So all angle bisectors of the quadrilateral $ABEF$ intersect at $I$, which means that it is circumscribable.