(Proof verification) For each $s \in S$, show that $\sum_{t \in Gs} \frac{1}{|Gt|}=1$

This is a lovely idea, but unfortunately the symbols have led us astray.

Consider the following functions and sums, where I will overdo the notation to help make the point: $$ f(s) = \sum_{t \in Gs} \frac{1}{|Gt|} = \sum_{\substack{t\in S \\ t \in Gs}} \frac{1}{|Gt|} $$ is a function of $s$ (not $t$, which is a dummy variable), and $$ g(s) = \sum_{s \in Gt}\frac{1}{|Gt|} = \sum_{\substack{t\in S \\ s \in Gt}} \frac{1}{|Gt|} $$ is again a function of $s$ (not $t$, which is a dummy variable); but $$ h(t) = \sum_{s \in Gt}\frac{1}{|Gt|} = \sum_{\substack{s\in S \\ s \in Gt}} \frac{1}{|Gt|} $$ is a function of $t$ (not $s$, which is a dummy variable), and $$ j(t) = \frac{1}{|Gt|}\sum_{s \in Gt}1 = \frac{1}{|Gt|}\sum_{\substack{s\in S \\ s \in Gt}} 1 $$ is again a function of $t$ (not $s$, which is a dummy variable).

You are correct that $f(s)=g(s)$, since $t\in Gs$ if and only if $s\in Gt$. You are also correct that $h(t)=j(t)$ (one can factor out from a sum an expression that's independent of the dummy variable) and that $j(t)=1$.

However, $g(s)$ and $h(t)$ are different functions (of different variables, even), and while it turns out that they happen to both equal the same constant, that fact has to be established by some argument. In the OP there was no such argument, and indeed the more concise notation in the two definitions (which is the same for both functions) subtly asserted that $g(s)=h(t)$ without proof (and without us noticing).