$\angle (BA,BC)=\angle BAC+\angle ACB=\angle BAC+\angle DAB=\angle DAC.$ Similarly, we have $\angle (BD,BA)=\angle DAC$ $\Longrightarrow$ $\angle DBC=2\angle DAC.$ Thus, $ED=EC$ and $\angle DEC=\angle DBC=2\angle DAC$ imply that $E$ is the circumcenter of $\triangle ACD.$ Since $KL$ is antiparallel to $CD$ with respect to the lines $AC,AD,$ then it follows that $KL$ is perpendicular to the A-circumdiameter $AE$ of $\triangle ACD,$ as desired.

Under inversion wrt $A$ suppose $B,C,D$ goes to $B',C',D'$. $AC'B'D'$ is a parallelogram. Suppose, $E$ goes to $E'$. Clearly, $\frac {CE'}{DE'}=\frac {AC}{AD}$. Since $E'$ lies on $\odot B'C'D'$, so it is the reflection of $A$ on $C'D'$. So $AE'\perp C'D'$. So $AE\perp KL$,since $KL$ is anti-parallel to $CD$.

$\angle KAL = \angle KAB + \angle BAL = \angle ACB + \angle ADB = \angle BDL + \angle ADB = \angle ADL$ $\Longrightarrow$ $\angle I_1 D L = \angle I_1 A L = 90^{\circ}.$ So $AD$ is the polar of $L$ wrt circle $(I_1)$ and $I_1L \perp AD$ (or AK), $I_2 K \perp AL.$ Follows $E$ is the orthocentre of $\triangle AKL.$ Hence, $AE \perp KL.$ [Moderator edit: LaTeXed]

it suffices to prove that A is the orthocenter of triangle EKL and by angle substitutions it's easy to obtain $AK\bot EL,AL\bot EK$.

$E$ is circumcenter tringle $ADC$

Observe that: $$\angle DAB = \angle ACB = \angle BDK = \alpha \quad \text{and} \quad \angle CAB = \angle ADB = \angle BCL =\beta$$ Thus $LC$ is tangent to $\mathcal K_2$ and $LD$ is tangent to $\mathcal K_1$ meaning that $LA=LC$ and $KA = KD$. Claim: Point $E$ is the circumcenter of $\odot(ADC)$ Proof: Just note that $DE=EC$, since $E$ is the midpoint of arc $EC$. On the other hand $\angle DAC = \alpha+\beta$, whereas $\angle LAC = \angle DAC = 180^{\circ} -2 \alpha -2\beta$. Thus $\angle DEC = 2(\alpha+\beta)$ implying that $E$ is a circumcenter as desired. To finish the problem observe that points $K, E, I_1$ all lie on the perpendicular bisector of the segment $AD$ and similarly points $L,E,I_2$ all lie on the perpendicular bisector of the segment $AC$. Therefore $E$ is the orthocenter of $\triangle AKL$ meaning that $AE \perp KL$ as desired.

littletush wrote: it suffices to prove that A is the orthocenter of triangle EKL and by angle substitutions it's easy to obtain $AK\bot EL,AL\bot EK$. Indeed, I am fairly sure that one does not have to go through justifying that $E$ is the circumcenter of $ACD$, only pure angle chasing suffices to obtain the orthocenter.