I was wondering whether one could solve a differential equation with initial value at $x_0\neq 0$. I think that the series must converge for $x=x_0$, otherwise the IVP wouldn't make sense. It seems as though one needs to know the value at which the series converges evaluated at $x_0$. However, it might be difficult to know this value when one can't express the series as a sum ($\sum$ notation) but rather as a sum of terms in the form $a_1+a_2x+\dotsc$.
So, could one solve an IVP not at $0$ if we have a strange summation? Is there any technique apart from finding the value at which the series converges?
There's several questions (it seems) involved here, some of which are unclear.
To answer your first question: yes, you can solve an IVP with conditions specified at $x_0\not=0$. You simply take the power series to be centered at $x_0$ and proceed as always to determine the coefficients $a_n$. $$ y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots\tag{1} $$ Note that this ansatz does not result in terms of the form $a_0+a_1x+a_2x^2+\cdots$ as you said. But this is a good thing since our initial condition(s) are specified at $x_0\not=0$.
Just like when $x_0=0$, your method will differ depending on whether $x_0$ is an ordinary point or singular point of the given equation.
As for a technique which determines the values of $x$ for which the series above converges, the answer again is "yes". You find the radius and interval of convergence of $(1)$ just like you did in calculus class (e.g., using the Ratio Test and checking the endpoints separately).