I feel like I overcomplicated this a lot, but here we go. Let $I$ be the circumcenter of triangle $DEF$ and $Q'$ be the isogonal conjugate of point $Q$. Furthermore, let $(\Gamma)$ be the circle $(AEF)$ and $(\Omega)$ be the circle $(ABC)$. The proof is structured in several Claims. Claim 1: Point $I$ is actually the incenter of triangle $ABC$ (surprise!) Proof: Note that $BF=BD$ and $IF=ID$, hence $BI \perp FD$, and similarly $CI \perp ED$. These two imply that $BI$ and $CI$ are the respective angle bisectors, as desired $\blacksquare$ Claim 2: Point $I$ belongs to circle $(\Gamma)$. Proof: Suppose not, and let the angle bisector of angle $A$ intersect $(\Gamma)$ at point $I'$. Then, $$\angle I'FE=\angle I'AE=\angle I'AF=\angle I'EF,$$implying that $I'E=I'F$, and so $I' \equiv I$, absurd $\blacksquare$ Claim 3: Redefine $Q$ as the second intersection of $AD$ and circle $(\Gamma)$. Then, $IQ,EF$ and $BC$ concur. Proof: Let $(\Gamma)$ intersect $BC$ again at point $Y \neq D$. Then, $$\angle IYD=\angle IDY=\angle IEC=\angle AFI=\angle AQI,$$and so quadrilateral $QIYD$ is cyclic. Hence, $IQ,EF$ and $BC$ concur at the radical center of circles $(\Gamma), (DEF)$ and $(QIYD)$ $\blacksquare$ Claim 4: Quadrilateral $PSQD$ is cyclic. That is, the definition of $Q$ we gave matches the original one. Proof: Note that $$\angle FSB=180^\circ-\angle C-\angle ASF=180^\circ-\angle C-\angle FEC=\angle A+\angle B-\angle FEC=\angle B-\angle AFE=\angle FPB,$$ and so $PSFB$ is cyclic. Thus, $$\angle SPD=\angle SFA=\angle SQA,$$ and so $PSQD$ is cyclic $\blacksquare$ We're in good shape. We may as well eliminate points $S,P$ and $Y$ now. Now we will bring point $Q'$ into play. Suppose that $AQ'$ and $AI$ intersect circle $(\Omega)$ at points $X$ and $M$ respectively. It suffices to show $IQ' \parallel XM$. Indeed, if so then we have $$\angle AQ'I=\angle AXM=\angle ACM,$$ and so $Q'$ belongs to a fixed circle (which will also be internally tangent to $(\Omega)$ at point $A$, but we don't care). So, here's our last Claim: Claim 5: $IQ' \parallel XM$. Proof: It suffices to show that $$\dfrac{AQ'}{Q'X}=\dfrac{AI}{IM}$$ Note that $$\dfrac{AQ'}{Q'X}=\dfrac{AB}{BX} \cdot \dfrac{\sin \angle ABQ'}{\sin \angle Q'BX}=\dfrac{\sin \angle C}{\sin \angle CAQ} \cdot \dfrac{\sin \angle QBC}{\sin \angle BQD},$$ where the angle equality $\angle Q'BX=\angle BQD$ holds because $\angle Q'BX=\angle Q'BC+\angle CBX=\angle ABQ+\angle BAQ=\angle BQD$. However, $\dfrac{\sin \angle C}{\sin \angle CAQ} \cdot \dfrac{\sin \angle QBC}{\sin \angle BQD}=\dfrac{AD}{DC} \cdot \dfrac{QD}{BD},$ and so we just have to show that $\dfrac{DQ \cdot DA}{DB \cdot DC}$ doesn't depend on the position of point $D$ (if we show this then we are basically done by taking $D$ to be the point where $AI$ meets $BC$, by harmonic ratios). We will use the linearity of power of a point, because we love bash. Let $(A)$ be the circle with center $A$ and radius, well, $0$. Suppose furthermore that $\dfrac{BD}{DC}=\lambda$. By trivial vector calculations, $\overrightarrow{D}=\dfrac{1}{\lambda+1}\overrightarrow{B}+\dfrac{\lambda}{\lambda+1}\overrightarrow{C}$. You can see what's coming, right? We have that ${\rm pow} (D,\Gamma,A)=\dfrac{1}{\lambda+1}{\rm pow} (B,\Gamma,A)+\dfrac{\lambda}{\lambda+1} {\rm pow} (C,\Gamma,A),$ and so $DQ \cdot DA-DA^2=\dfrac{1}{\lambda+1}(\dfrac{ac\lambda}{\lambda+1}-c^2)+\dfrac{\lambda}{\lambda+1}(\dfrac{ab}{\lambda+1}-b^2)$. Now we use Stewart's to compute $DA^2$. We get $DA^2=\dfrac{c^2}{\lambda1+}+\dfrac{b^2\lambda}{\lambda+1}-\dfrac{a^2\lambda}{(\lambda+1)^2}$, and now putting this back into the above relation we will eventually get that $$DQ \cdot DA=\dfrac{c+b-a}{a} \cdot \dfrac{a^2\lambda}{(\lambda+1)^2}=\dfrac{c+b-a}{a} \cdot DB \cdot DC,$$hence $\dfrac{DQ \cdot DA}{DB \cdot DC}=\dfrac{c+b-a}{a},$ as desired! $\blacksquare$

Here is the official solution: $I \in (AEF)$. $S,D,L$ are collinear. $SI \cap EF=J$. Spiral similarity taking $EF$ to $BC$, takes $J$ to $D$. Thus $J \in (SPD)$. If $Q=IP \cap (AEF)$ then $IJ \cdot IS =IQ \cdot IP=ID^2$ and $\angle IQA= \angle AFI=\angle IDC=\angle PQD \implies Q = (PSD) \cap AD$. Let $R=(AIZ) \cap AW$. $X$ is the intersection of $A$-angle bisector and $BC$. $(ARC) \cap BC=G$. $D' \in (DEF)$ is the reflection of $D$ across $T$. $U=AW \cap (DEF)$. $\angle UD'Y=\frac{1}{2} \angle DIU=\angle XIU$. Also $\angle IRU=\angle ABL=\angle IXC=180- \angle IXD=180- \angle IXU$. Thus $IRUD'X$ is cyclic. $\textbf{Claim:}$ $\triangle RXG \sim \triangle QFA$ $\angle IPD= \angle IDQ= \angle IUA= \angle ID'R$ and $\angle IRD'=\angle IXP \implies \triangle RID' \sim \triangle XIP$ $ \implies \angle RXG=\angle RID' =\angle PIX=\angle AFQ$. $\angle RGC=\angle RAC=\angle QAF$. $\square$ $\textbf{Claim:}$ $\triangle RWC \sim \triangle XBF$ $\angle XBF=\angle RWC$. $\frac{RW}{WC}=\frac{WA \cdot LI}{WC \cdot LA}=\frac{AB \cdot LI}{BD \cdot LA}=\frac{BX}{BF}$.$\square$ To finish the proof we have $\angle FXB=\angle WRC=\angle AGB \implies XF \parallel AG $$\implies \frac{BF}{FA}=\frac{BX}{XG} \implies \triangle RXG \cap B \sim QFA \cap B $ $\implies \angle ABQ= \angle RBC$. Similarly $\angle ACR= \angle BCQ$.

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