Let $X$ be the second intersection point of $\Gamma$ with $IA'$, where $A'$ is the antipodal of $A$ with respect to $\Gamma$. We will prove that the points $X, S, I$ are collinear.

We have $\widehat{IXA} = 90^\circ$, so $X$ belongs to the circle of diameter $AI$. The same holds for the points $E$ and $F$, thus $X$ is on the circumcircle of $AEIF$. Therefore $\widehat{AFX} = \widehat{AEX}$. However, $\widehat{ACX} = \widehat{ABX}$, and as a result the triangles $BFX, CEX$ are similar. It follows that,
$$\frac{XF}{XE} = \frac{BF}{CE} = \frac{BD}{CD} \qquad (1).$$
Since the intersection of $EF$ with $BC$ is the isogonal conjugate of $B$, and $\widehat{SDF} = 90^\circ$, we get that $SD$ is the bisector of $\widehat{BSC}$, thus
$$\frac{BS}{CS} = \frac{BD}{CD} \qquad (2).$$
Finally, since $\widehat{AFE} = \widehat{AEF}$, and $\widehat{FSB} = \widehat{CSE}$, the triangles $BFS, CES$ are similar, so
$$\frac{BS}{CS} = \frac{SF}{SE} \qquad (3).$$
From (1), (2), (3) we get $\frac{XF}{XE} = \frac{BD}{CD} = \frac{BS}{CS} = \frac{SF}{SE}$, which gives us that $XS$ is the bisector of $\widehat{EXF}$. Since $I$ is the midpoint of the arc $EF$ at ($AEIF$), we have that $XI$ is the bisector of $\widehat{EXF}$, so $X, S, I$ are collinear

fig. 5

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Let $M, N$ the midpoints of $FD, DE$, then $\widehat{DFE} = \widehat{DEC}$, and the right-angled triangles DFS, DCN are similar, so:
$$\frac{FS}{DF} = \frac{EN}{DC} \qquad (4)$$
Similarly, from the similar triangles DES, BDM we get:
$$\frac{SE}{ED} = \frac{MD}{BD} \qquad (5)$$
The relations (4) and (5) give
$$\frac{FS}{SE} = \frac{BD}{CD} \qquad (6)$$
The last one with the help of (1) gives that XS is the bisector of $\widehat{EXF}$. Since I is the midpoint of the arc EF at (AEIF), we have that XI is the bisector of $\widehat{EXF}$, so X, S, I are collinear.