Let $P$ be the second intersection of the circumcircles of triangles $TCF$ and $TDE$. Because the quadrilateral $PEDT$ is cyclic, $\angle PET = \angle PDT$, or
$$\angle PEF = \angle PDC. \quad (*)$$
Because the quadrilateral $PFCT$ is cyclic,
$$\angle PFE = \angle PFT = \angle PCT = \angle PCD. \quad (**)$$
By equations () and (*), it follows that triangle $PEF$ is similar to triangle $PDC$. Hence $\angle FPE = \angle CPD$ and $PF/PE = PC/PD$. Note also that $\angle FPC = \angle FPE + \angle EPC = \angle CPD + \angle EPC = \angle EPD$. Thus, triangle $EPD$ is similar to triangle $FPC$. Another way to say this is that there is a spiral similarity centered at $P$ that sends triangle $PFE$ to triangle $PCD$, which implies that there is also a spiral similarity, centered at $P$, that sends triangle $PFC$ to triangle $PED$, and vice versa. In terms of complex numbers, this amounts to saying that
$$\frac{D-P}{E-P} = \frac{C-P}{F-P} \implies \frac{E-P}{F-P} = \frac{D-P}{C-P}.$$

Because $AE/ED = BF/FC$, points $A$ and $B$ are obtained by extending corresponding segments of two similar triangles $PED$ and $PFC$, namely, $DE$ and $CF$, by the identical proportion. We conclude that triangle $PDA$ is similar to triangle $PCB$, implying that triangle $PAE$ is similar to triangle $PBF$. Therefore, as shown before, we can establish the similarity between triangles $PBA$ and $PFE$, implying that
$$\angle PBS = \angle PBA = \angle PFE = \angle PFS \quad \text{and} \quad \angle PAB = \angle PEF.$$

We will give a solution using complex coordinates. The first step is the following lemma.

Lemma Suppose $s$ and $t$ are real numbers and $x$, $y$ and $z$ are complex. The circle in the complex plane passing through $x$, $x + ty$ and $x + (s + t)z$ also passes through the point $x + syz/(y - z)$, independent of $t$.

Proof: Four points $z_1, z_2, z_3$ and $z_4$ in the complex plane lie on a circle if and only if the cross-ratio
$$cr(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}$$
is real. Since we compute
$$cr(x, x + ty, x + (s + t)z, x + syz/(y - z)) = \frac{s + t}{s}$$
the given points are on a circle.

Lay down complex coordinates with $S = 0$ and $E$ and $F$ on the positive real axis. Then there are real $r_1, r_2$ and $R$ with $B = r_1A$, $F = r_2E$ and $D = E + R(A - E)$ and hence $AE/ED = BF/FC$ gives
$$C = F + R(B - F) = r_2(1 - R)E + r_1RA.$$
The line $CD$ consists of all points of the form $sC + (1-s)D$ for real $s$. Since $T$ lies on this line and has zero imaginary part, we see from $\text{Im}(sC + (1-s)D) = (sr_1R + (1-s)R)\text{Im}(A)$ that it corresponds to $s = -1/(r_1 - 1)$. Thus
$$T = \frac{r_1D - C}{r_1 - 1} = \frac{(r_2 - r_1)(R - 1)E}{r_1 - 1}.$$
Apply the lemma with $x = E$, $y = A - E$, $z = (r_2 - r_1)E/(r_1 - 1)$, and $s = (r_2 - 1)(r_1 - r_2)$. Setting $t = 1$ gives
$$(x, x + y, x + (s + 1)z) = (E, A, S = 0)$$
and setting $t = R$ gives
$$(x, x + Ry, x + (s + R)z) = (E, D, T).$$
Therefore the circumcircles to $SAE$ and $TDE$ meet at
$$x + \frac{syz}{y - z} = \frac{AE(r_1 - r_2)}{(1 - r_1)E - (1 - r_2)A} = \frac{AF - BE}{A + F - B - E}.$$
This last expression is invariant under simultaneously interchanging $A$ and $B$ and interchanging $E$ and $F$. Therefore it is also the intersection of the circumcircles of $SBF$ and $TCF$.