Let $\omega_1$ and $\omega_2$ be the incircles and $O_1$ and $O_2$ the incenters of the quadrilaterals $ABED$ and $AECD$ respectively. A point $F$ with the stated property exists only if $\omega_1$ and $\omega_2$ are also the incircles of the quadrilaterals $ABCF$ and $BCDF$.

Let the tangents from $B$ to $\omega_2$ and from $C$ to $\omega_1$ (other than $BC$) meet $AD$ at $F_1$ and $F_2$ respectively. We need to prove that $F_1 = F_2$ if and only if $AB \parallel CD$.

Lemma. The circles $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ are inscribed in an angle with vertex $O$. The points $P, S$ on one side of the angle and $Q, R$ on the other side are such that $\omega_1$ is the incircle of the triangle $PQO$, and $\omega_2$ is the excircle of the triangle $RSO$ opposite to $O$. Denote $p = OO_1 \cdot OO_2$. Then exactly one of the following relations holds:
$$OP \cdot OR < p < OQ \cdot OS, \quad OP \cdot OR > p > OQ \cdot OS, \quad OP \cdot OR = p = OQ \cdot OS.$$
Proof. Denote $\angle OPO_1 = u$, $\angle OQO_1 = v$, $\angle OO_2R = x$, $\angle OO_2S = y$, $\angle POQ = 2\varphi$. Because $PO_1$, $QO_1$, $RO_2$, $SO_2$ are internal or external bisectors in the triangles $PQO$ and $RSO$, we have
$$\begin{equation*} u + v = x + y \left(= 90^\circ - \varphi\right). \tag{1} \end{equation*}$$

By the law of sines
$$\frac{OP}{OO_1} = \frac{\sin(u+\varphi)}{\sin u} \quad \text{ and } \quad \frac{OO_2}{OR} = \frac{\sin(x+\varphi)}{\sin x}.$$
Therefore, since $x, u$ and $\varphi$ are acute,
$OP \cdot OR \geq p \Leftrightarrow \frac{OP}{OO_1} \geq \frac{OO_2}{OR} \Leftrightarrow \sin x \sin(u+\varphi) \geq \sin u \sin(x+\varphi) \Leftrightarrow \sin(x-u) \geq 0 \Leftrightarrow x \geq u$.
Thus $OP \cdot OR \geq p$ is equivalent to $x \geq u$, with $OP \cdot OR = p$ if and only if $x = u$.
Analogously, $p \geq OQ \cdot OS$ is equivalent to $v \geq y$, with $p = OQ \cdot OS$ if and only if $v = y$. On the other hand $x \geq u$ and $v \geq y$ are equivalent by (1), with $x = u$ if and only if $v = y$. The conclusion of the lemma follows from here.

Going back to the problem, apply the lemma to the quadruples $\{B, E, D, F_1\}, \{A, B, C, D\}$ and $\{A, E, C, F_2\}$. Assuming $OE \cdot OF_1 > p$, we obtain
$$OE \cdot OF_1 > p \Rightarrow OB \cdot OD < p \Rightarrow OA \cdot OC > p \Rightarrow OE \cdot OF_2 < p.$$
In other words, $OE \cdot OF_1 > p$ implies
$$OB \cdot OD < p < OA \cdot OC \text{ and } OE \cdot OF_1 > p > OE \cdot OF_2.$$
Similarly, $OE \cdot OF_1 < p$ implies
$$OB \cdot OD > p > OA \cdot OC \text{ and } OE \cdot OF_1 < p < OE \cdot OF_2.$$
In these cases $F_1 \neq F_2$ and $OB \cdot OD \neq OA \cdot OC$, so the lines $AB$ and $CD$ are not parallel.

There remains the case $OE \cdot OF_1 = p$. Here the lemma leads to $OB \cdot OD = p = OA \cdot OC$ and $OE \cdot OF_1 = p = OE \cdot OF_2$. Therefore $F_1 = F_2$ and $AB \parallel CD$.