Let $\alpha=\angle C A B$, $\beta=\angle A B C$ and $\gamma=\angle B C A$. The line $A J$ is the bisector of $\angle C A B$, so $\angle J A K=\angle J A L=\frac{\alpha}{2}$. By $\angle A K J=\angle A L J=90^{\circ}$ the points $K$ and $L$ lie on the circle $\omega$ with diameter $A J$.

The triangle $K B M$ is isosceles as $B K$ and $B M$ are tangents to the excircle. Since $B J$ is the bisector of $\angle K B M$, we have $\angle M B J=90^{\circ}-\frac{\beta}{2}$ and $\angle B M K=\frac{\beta}{2}$. Likewise $\angle M C J=90^{\circ}-\frac{\gamma}{2}$ and $\angle C M L=\frac{\gamma}{2}$. Also $\angle B M F=\angle C M L$, therefore
$$\angle L F J=\angle M B J-\angle B M F=\left(90^{\circ}-\frac{\beta}{2}\right)-\frac{\gamma}{2}=\frac{\alpha}{2}=\angle L A J .$$
Hence $F$ lies on the circle $\omega$. (By the angle computation, $F$ and $A$ are on the same side of $B C$.) Analogously, $G$ also lies on $\omega$. Since $A J$ is a diameter of $\omega$, we obtain $\angle A F J=\angle A G J=90^{\circ}$.

The lines $A B$ and $B C$ are symmetric with respect to the external bisector $B F$. Because $A F \perp B F$ and $K M \perp B F$, the segments $S M$ and $A K$ are symmetric with respect to $B F$, hence $S M=A K$. By symmetry $T M=A L$. Since $A K$ and $A L$ are equal as tangents to the excircle, it follows that $S M=T M$, and the proof is complete.