I do not understand the step where A(x) - 1 - 2xA(x) = 0. I am guessing that it has something to do with the powers of n being different in each summation (n vs n-1), but I do not understand how that is dealt with by simply subtracting 1.
Please explain why subtracting 1 would solve the different powers of n problem. If that is not what solves that problem, please explain why we subtract 1, and then why the different powers of n do not matter.
Assuming $A$ is defined by $A(x) = \sum_{n=0}^\infty a_n x^n$ (note the index starts at $0$), we have $$ \sum_{n\geq 1} a_n x^n = \sum_{n\geq 0} a_n x^n - a_0 x^0 = A(x) -a_0 = A(x) - 1 $$ since $a_0=1$.
Moreover, $\sum_{n\geq 1} a_{n-1} x^{n-1} = \sum_{n\geq 0} a_{n} x^{n} = A(x)$ by a change of indices, so $$ \sum_{n\geq 1} a_n x^n - 2x \sum_{n\geq 1} a_{n-1} x^{n-1} = (A(x) - 1) - 2x A(x) = A(x) - 1 - 2x A(x) $$ as stated.