How to obtain the asymptotics of Legendre polynomials directly from their generating function

As described in Analytic Combinatorics by Flajolet and Sedgewick, page 4, the pole $t_0$ of the generating function $F(t)$ of smallest absolute value governs the exponential asymptotics $P_n\sim (1/t_0)^n$. In this case $t_0=x-\sqrt{x^2-1}$, hence $P_n\sim (x+\sqrt{x^2-1})^n$.

To obtain the subexponential factor one expands $F(t)$ around $t_0$, $$F(t)\simeq 2^{-1/2}(x^2-1)^{-1/4}t_0^{-1/2}(1-t/t_0)^{-1/2}$$ $$=2^{-1/2}(x^2-1)^{-1/4}t_0^{-1/2}\sum_{n = 0}^{\infty}\frac{(2n - 1)!!}{2^n n!}(t/t_0)^n$$ $$\simeq (2\pi )^{-1/2}(x^2-1)^{-1/4}\sum_{n}n^{-1/2}(1/t_0)^{n+1/2}\,t^n,$$ which gives precisely the large-$n$ asymptotics for $P_n$ quoted in the OP.

You may write $2x=a+1/a$ for certain $a$, $|a|>1$ (I guess you mean $|x|>1$), then $$ \frac1{\sqrt{1-2tx+t^2}}= \frac1{\sqrt{(1-at)(1-a^{-1}t)}}\\= \sum (-1)^n{-1/2\choose n}a^nt^n\cdot \sum (-1)^n{-1/2\choose n}a^{-n}t^n\\ :=\sum c_na^nt^n\cdot \sum c_na^{-n}t^n,\quad c_n=(-1)^n{-1/2\choose n}=\frac{1\cdot 3\cdot\ldots \cdot(2n-1)}{2\cdot 4\cdot \ldots \cdot (2n)}\sim \frac1{\sqrt{\pi n}} $$ (see the last relation in "Additional identities" section here, or deduce from the Wallis formula). Thus $$P_n(x)=\sum_{k=0}^n c_{n-k}c_k a^{n-2k}.$$ This gives (only small values of $k$ matter) the asymptotics $$P_n(x)a^{-n}\sqrt{\pi n}\to_{(*)} \sum_{k=0}^\infty c_k a^{-2k}=\frac1{\sqrt{1-a^{-2}}}$$ that is equivalent to what you ask for.

More details for $(*)$. We have $$ P_n(x)a^{-n}\sqrt{\pi n}-\sum_{k=0}^\infty c_k a^{-2k}=\sum \alpha_kc_ka^{-2k}, $$ where $$\alpha_k=\begin{cases}\sqrt{\pi n}c_{n-k}-1,& k\leqslant n\\ -1,&k>n.\end{cases}$$ For any fixed $k$ we have $\alpha_k\to 0$ when $n\to \infty$. On the other hand, $-1\leqslant \alpha_k\leqslant 100k$, say (for $k\geqslant n/2$ use the bound $c_{n-k}\leqslant 1$; for $k<n/2$ use the bound $c_{n-k}\leqslant c_{\lceil n/2\rceil}\leqslant 1/\sqrt{\pi n/2}$), thus the series $\sum \alpha_k c_k a^{-2k}$ is dominated by an absolutely convergent series $\sum 100 kc_ka^{-2k}$ and its sum goes to 0 by Dominated Convergence Theorem.)