Always note that for solving ODEs by power series or frobenius method approach, only the ODEs of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ can find all of the coefficients most nicely, since they only involves two terms recurrence relations with variable coefficients.
$y''+y'+xy=0$ is obviously not belongs to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ , so it should be better try to convert the ODE to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ first by some variable transformations.
Let $y=e^{ax}u$ ,
Then $y'=e^{ax}u'+ae^{ax}u$
$y''=e^{ax}u''+ae^{ax}u'+ae^{ax}u'+a^2e^{ax}u=e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u$
$\therefore e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u+e^{ax}u'+ae^{ax}u+xe^{ax}u=0$
$e^{ax}u''+(2a+1)e^{ax}u'+(x+a^2+a)e^{ax}u=0$
$u''+(2a+1)u'+(x+a^2+a)u=0$
Choose $a=-\dfrac{1}{2}$ , the ODE becomes $u''+\left(x-\dfrac{1}{4}\right)u=0$ , which belongs to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$
Let $u=\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^n$ ,
Then $u'=\sum\limits_{n=0}^\infty na_n\left(x-\dfrac{1}{4}\right)^{n-1}=\sum\limits_{n=1}^\infty na_n\left(x-\dfrac{1}{4}\right)^{n-1}$
$u''=\sum\limits_{n=1}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}=\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}$
$\therefore\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\left(x-\dfrac{1}{4}\right)\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^{n+1}=0$
$\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\sum\limits_{n=3}^\infty a_{n-3}\left(x-\dfrac{1}{4}\right)^{n-2}=0$
$2a_2+\sum\limits_{n=3}^\infty(n(n-1)a_n+a_{n-3})\left(x-\dfrac{1}{4}\right)^{n-2}=0$
$\therefore\begin{cases}2a_2=0\\n(n-1)a_n+a_{n-3}=0\end{cases}$
$\begin{cases}a_2=0\\a_n=-\dfrac{a_{n-3}}{n(n-1)}\end{cases}$
$\therefore\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^na_0}{(2\times3)(5\times6)(8\times9)......((3n-1)3n)}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^na_1}{(3\times4)(6\times7)(9\times10)......(3n(3n+1))}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^n(4\times7\times10\times......(3n+1))a_0}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n(2\times5\times8\times......(3n-1))a_1}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k+1)\biggr)a_0}{(3n+1)!}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)a_1}{(3n+1)!}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}a_{3n}=\dfrac{(-1)^n\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)a_0}{(3n+1)!}\forall n\in\mathbb{Z}^*\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)a_1}{(3n+1)!}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\therefore y=C_1e^{-\frac{1}{2}\left(x-\frac{1}{4}\right)}\sum\limits_{n=0}^\infty\dfrac{\biggl((-1)^n\prod\limits_{k=0}^n(3k+1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n}}{(3n+1)!}+C_2e^{-\frac{1}{2}\left(x-\frac{1}{4}\right)}\biggl(x-\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n+1}}{(3n+1)!}\biggr)=C_1\biggl(\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{2^nn!}\left(x-\dfrac{1}{4}\right)^n\biggr)\sum\limits_{n=0}^\infty\dfrac{\biggl((-1)^n\prod\limits_{k=0}^n(3k+1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n}}{(3n+1)!}+C_2\biggl(\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{2^nn!}\left(x-\dfrac{1}{4}\right)^n\biggr)\biggl(x-\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n+1}}{(3n+1)!}\biggr)$
Expand it you will get the power series solution centered at $\dfrac{1}{4}$