The Theorem and the proof are as follows, where $L_{2n}$ is the number of steps of the walk which were not within the negative half-line, and $T_0$ is the first time of the walk revisit the position $0$. And the reference to $(1b)$ of the proof is $$ P_0(s)=\sum_{n=0}^{\infty}P(S_n=0)s^n=\sum_{n=0}^{\infty}p_0(n)s^n=(1-4pqs^2)^{-\frac{1}{2}}. $$ I understand the proof except fail to verify the last equation. Why $$ \frac{2\big[\sqrt{1-s^2t^2}-\sqrt{1-t^2}\big]}{t^2(1-s^2)}=\sum_{n=0}^{\infty}t^{2n}\mathbb{P}(S_{2n}=0)\bigg(\frac{1-s^{2n+2}}{(n+1)(1-s^2)}\bigg)? $$ I replaced $\sqrt{1-s^2t^2}\;$ by $\;(1-s^2t^2)\cdot P_0(st)\;$ and $\sqrt{1-t^2}\;$ likewise, but couldn't get the same result.
The following content is in Page 169 of Grimmett's Probability and Random Processes
I don't know much about what you learn but I think you have overthought this problem and chosen a not true approach.
So you see if we have to deal with an analytic function $f(x)$ with the convergence radius $1$ : $$ f(x) := \sum_{n \ge 0} a_nx^n$$ in order to obtain a compact form of the analytic expansion as you wrote at the beginning of your post, we can integrate $f$ and do some substitution and subtraction :
Integrating from $0$ to $t$ $$ F(t) := \int_{0}^t f(x)dx = \sum_{n \ge 0} \dfrac{a_n}{n+1}t^{n+1} $$
Substitution and subtraction: $$ \dfrac{F(t^2)-F(t^2s^2)}{t^2(1-s^2)} = \sum_{n \ge 0} \dfrac{a_n}{n+1} t^{2n} \left( \dfrac{ 1-s^{2n+2}}{1-s^2} \right)$$
By letting $a_n= C_n$; $f(x)= (1-x^2)^{-1/2} $, you get what you want.