So this is my perspective on the problem, and I do think the other question seems to provide a reasonable proof. However, I'll take a different approach since maybe the issue is how the initial equation is arrived at. However, this will require linear algebra and this may or may not be satisfying. I do think, to an extent, there is a level of handwaving that is necessary if very little background is assumed. With that disclaimer out of the way...
Suppose there was a really large room, with a large number $N \gg 0$ of these games being executed, turn-by-turn, in parallel. Consider the sequence of proportions of games that continue to run in a given turn as given by
$$P_1, P_2, \ldots, P_k, \ldots,$$
the proportion of games where the player with Black has won given by,
$$B_1, B_2, \ldots, B_k, \ldots,$$
and finally the proportion of games where the player with White has won given by,
$$W_1, W_2, \ldots, W_k, \ldots.$$
Effectively, $B_k$ is the probability of you picking any given game and seeing that the player playing Black had won in the time history of the game prior to turn $k$. Based on the problem description we know that these proportions observe the following recursion,
$$
\begin{aligned}
P_{k+1} &= p q P_k\\
B_{k+1} &= p P_k + B_k\\
W_{k+1} &= q^2 P_k + W_k
\end{aligned}
$$
Explicitly this encodes that,
- the proportion of running games $P_k$ proceeding to the next turn with no winners $P_{k+1}$ is $p q$.
- $p$ percent of those currently running games $P_k$ result in the player playing Black winning.
- $q^2$ percent of those currently running games $P_k$ result in the player playing White winning.
- Those games with winners, i.e. $W_k$ and $B_k$, proceed to stay in that state.
I think this law is clear from the rules of the game, and doesn't rely on anything more. Under the assumption that $q = 1 - p$, you can check that the sum of these proportions stay equal to $1$ so this recursion makes sense. Use a little linear algebra to write this as,
$$
\begin{pmatrix} P_{k+1} \\ B_{k+1} \\ W_{k+1}\end{pmatrix}
=
\begin{pmatrix} p q & 0 & 0 \\ p & 1 & 0 \\ q^2 & 0 & 1 \end{pmatrix}
\begin{pmatrix} P_k \\ B_{k} \\ W_{k}\end{pmatrix}
$$
Consider the following question:
If it exists and is non-zero, what is the proportion of game where the player playing Black wins as $k\to\infty$?
That is, what is $B_k$ tend towards as $k\to\infty$? This is simply a discrete recursion and we have to verify (1) it converges, and (2) what it converges to. To aid in this, it helps to find a coordinate change $T$ where the matrix in question becomes diagonal. Define,
$$T = \begin{pmatrix} 1 - pq & 0 & 0\\ -p & 1 & 0 \\ -q^2 & 0 & 1\end{pmatrix}$$
and write
$$Z_k = T^{-1} \begin{pmatrix} P_k \\ B_k \\ W_k \end{pmatrix}$$
Verify that in these new coordinates the previous discrete recursion transforms into,
$$Z_{k+1} = \begin{pmatrix} p q & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} Z_k$$
We can find a closed-form solution to this recursion and clearly the first component of $Z_k$ vanishes as $k\to\infty$ (since $pq < 1$) while the other two components stay constant. As a result, the system converges as $k\to\infty.$ Because the first component of $Z_k$ vanishes as $k \to\infty,$ the last two components of $Z_k$ determine $B_k$ and $W_k$ in the limit. From this fact, deduce that,
$$Z_\infty^2 = B_\infty,\quad Z_\infty^3 = W_\infty,$$
where the superscript indicates the component of that vector and I'm abusing the subscript notation to indicate the limiting value as $k\to\infty.$ Technically, we could also just find the limiting behaviour of $Z_k$ and then use the transformation to figure out what $B_k$ and $W_k$ will tend to. Here I will take the shortcut.
So as long as we can figure out how the initial condition,
$$\begin{pmatrix} P_0 \\ B_0 \\ W_0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$
is written in $Z$ coordinates, we can compute what $B_\infty$ and $W_\infty$ will be: it will just be the last two components of $Z_0$. Ok, so that means we have to solve the equation,
$$ T Z_0 = \begin{pmatrix} P_0 \\ B_0 \\ W_0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$
If you run through the algebra, you will find that the first two equations in this system appear as,
$$\begin{aligned}
1 &= (1 - p q) Z_0^1\\
0 &= (-p) Z_0^1 + Z_0^2
\end{aligned}$$
Multiply the first equation by $p$ and eliminate $Z_0^1$ to find,
$$
p = -p q Z_0^2 + Z_0^2
$$
You should recognize this equation immediately. Recalling that $B_\infty = Z_0^2,$
$$
p = -p q B_\infty + B_\infty
$$
arriving at the critical equation. Solving it yields an answer to aforementioned question. Then it is a matter of arguing why we would consider this "probability of picking a random game and seeing the player playing Black winning" as a measure of "the probability of the player in Black winning".
Note this procedure relies on the initial intuition for the problem setup after which the rest is purely linear algebraic. Here the problem is simple since there is only three game states (one playing, two termination states). You could also do the exact same derivation with four play states or more states if you want, but the result should be same regardless. This approach builds into Markov Theory, and applies as long as you can model it with a Markov Model. It also can generalize to countable number of game states, but one has to be a bit more careful then.
