Solution:

Use complex coordinates and the standard convention where $O$ is the origin of the complex plane, and $\gamma$ is the unit circle; as usual, a lower case Roman letter denotes the complex coordinate of the point denoted by the corresponding upper case Roman letter.

Begin by expressing $t$ in terms of $p, q$ and their complex conjugates. Since $p'\bar{p} = 1$ and $q'\bar{q} = 1$, the equations of the lines $PQ'$ and $QP'$ are
$$\bar{q}(1 - \bar{p}q)z - q(1 - p\bar{q})\bar{z} + \bar{p}q - p\bar{q} = 0 \quad \text{and} \quad \bar{p}(1 - \bar{q}p)z - p(1 - q\bar{p})\bar{z} + \bar{q}p - q\bar{p} = 0,$$
so the two cross at
$$t = \frac{p + q - pq(\bar{p} + \bar{q})}{(1 - p\bar{p}q\bar{q})^2}.$$

Next, we turn to isogonal conjugates in the triangle $ABC$. Two points $U$ and $V$ in the plane $ABC$, not lying on $\gamma$, are isogonally conjugate in the triangle $ABC$ if and only if $u + v + abc\bar{u}\bar{v} = a + b + c$; for completeness, a proof of this fact is provided at the end of the solution. In particular, $p + q + abc\bar{p}\bar{q} = a + b + c$.

Now, $r = p + q$, and the center of the nine-point circle has complex coordinate $(a+b+c)/2$, so $s = a+b+c-p-q = abc\bar{p}\bar{q}$; since $s\bar{s} = p\bar{p}q\bar{q} < 1$, the point $S$ does not lie on $\gamma$.

Let $U$ be the isogonal conjugate of $S$ in the triangle $ABC$ to write $s+u+abc\bar{s}\bar{u} = a+b+c$, and refer to the above expressions of $s$ to get $u+pq\bar{u} = p+q$. Elimination of $\bar{u}$ from the latter and its complex conjugate yields
$$u = \frac{p+q-pq(\bar{p}+\bar{q})}{1-p\bar{p}q\bar{q}} = t,$$
and the conclusion follows.

For completeness, we show that two points $U$ and $V$ in the plane $ABC$, not lying on $\gamma$, are isogonally conjugate in the triangle $ABC$ if and only if $u + v + abc\bar{u}\bar{v} = a + b + c$.

If $U$ and $V$ are isogonally conjugate in the triangle $ABC$, isogonality at $A$ implies that the product of $(u-a)/(b-a)$ and $(v-a)/(c-a)$ is real, so it is equal to its complex conjugate. Alternatively, but equivalently, $(u-a)(v-a) = a^2bc(\bar{u} - \bar{a})(\bar{v} - \bar{a})$, so $uv - a(u+v) + a^2 = a^2bc\bar{u}\bar{v} - abc(\bar{u} + \bar{v}) + bc$. Similarly, isogonality at $B$ yields $uv - b(u+v) + b^2 = ab^2c\bar{u}\bar{v} - abc(\bar{u} + \bar{v}) + ca$. Subtract the two, factor $a-b$ out and rearrange terms to get the desired relation.

Conversely, let $W$ be the isogonal conjugate of $U$ in the triangle $ABC$ to write $u+w+abc\bar{u}\bar{w} = a+b+c$. Elimination of $u$ from the latter and $u+v+abc\bar{u}\bar{v} = a+b+c$ yields $v - w + abc\bar{u}(\bar{v} - \bar{w}) = 0$, and elimination of $\bar{v} - \bar{w}$ from the latter and its complex conjugate yields $(1 - u\bar{u})(v - w) = 0$. Since $U$ does not lie on $\gamma$, the first factor is different from zero, so $v = w$. This establishes the converse and completes the proof.