How to combine the four Theorems in order to prove the statement?

Let $Q(s) = \frac{1-\tau(s)}{1-s}$. It is not hard to check that the coefficient $q_n$ of $s^n$ in $Q(s)$ is $$ q_n = 1-\sum_{m=0}^n \tau_m = \Pr[\mathfrak{T}_1 > n] \sim \frac{1}{\sqrt{\pi}} e^{-c} \frac{1}{\sqrt{n}}, $$ using Theorem 1. Clearly $q_n$ is monotonic. Taking $p = 1/2$ and using $\Gamma(1/2) = \sqrt{\pi}$, this shows that $$ q_n \sim \frac{1}{\Gamma(\pi)} n^{p-1} L(n), \qquad L(n) = e^{-c}. $$ Clearly $L(n)$ is slowly varying at infinity (it is constant), so Theorem 3 shows that as $s \to 1-$, $$ \frac{1-\tau(s)}{1-s} \sim \frac{e^{-c}}{\sqrt{1-s}}. $$ Theorem 2 implies that as $s \to 1-$, $$ p(s) = \frac{1}{1-s} \frac{1-s}{1-\tau(s)} \sim \frac{e^c}{\sqrt{1-s}}. $$ Applying Theorem 3 again for $p(s)$, $p=1/2$ and $L(n) = e^c$ (clearly $p_n$ is monotonic), we deduce that $$ p_n \sim \frac{1}{\sqrt{\pi}} n^{-1/2} e^c. $$

Theorem 4 implies that $$ p_n \sim \frac{\sigma}{\sqrt{2\pi}} n^{-1/2} \mathbb{E}(S_N)^{-1} = \sigma ((2\pi n)^{1/2} \mathbb{E}(S_N))^{-1}, $$ which is your statement.