$$ \frac{y''}{y'}+y' = f(x) $$
I set the following to be true:
$$ y = \sum_{n=0}^{\infty} a_n x^n $$
$$ f(x) = \sum_{n=0}^{\infty} b_nx^n $$
Therefore:
$$ y'' = y'(f(x)-y') $$
$$ \sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n] $$
$$ \sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n \sum_{k=0}^n (k+1)a_{k+1}] $$ Therefore $$ (n+1)(n+2)a_{n+2} = (b_n - (n+1)a_{n+1}) \sum_{k=0}^n (k+1)a_{k+1} $$
$$ b_n = \frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_k} + (n+1)a_{n+1} $$
$$ y = \sum_{n=0}^{\infty} (\frac{(n+2)(n+1)a_{n+2}}{\sum_{k=1}^{n+1} ka_{k}} + (n+1)a_{n+1})x^n $$
Now if i set $a_n = 1$
$$ y = \sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=0}^\infty (\frac{(n+2)(n+1)}{\sum_{k=1}^{n+1} k}x^n $$
$$ \sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2} $$
$$ y = \frac{1}{(x-1)^2} + \sum_{n=0}^\infty 2x^n $$
$$ \sum_{n=0}^\infty 2x^n = \frac{2}{1-x} $$
Therefore
$$ f(x) = \frac{2x-1}{(x-1)^2} $$
$$ y = \frac{1}{1-x} $$
Therefore
$$ \frac{-2(1-x)^2}{(1-x)^3}+\frac{1}{(1-x)^2} = \frac{2x-1}{(1-x)^2} $$
Which is true.
Your problem is that you took the terms involving $b_n$ out of the summation. Here is my take.
I'll start at $\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]$ but use different indices for the summations on the right.
$\begin{array}\\ \sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n &= \sum_{i=0}^{\infty} (i+1)a_{i+1} x^i \times \sum_{j=0}^{\infty} (b_j - (j+1)a_{j+1})x^j\\ &= \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} (i+1)a_{i+1} (b_j - (j+1)a_{j+1})x^{i+j}\\ &= \sum_{n=0}^{\infty}\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})x^{n}\\ &= \sum_{n=0}^{\infty}x^{n}\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})\\ \end{array} $
From this,
$\begin{array}\ (n+1)(n+2)a_{n+2} &=\sum_{j=0}^{n} (n-j+1)a_{n-j+1} (b_j - (j+1)a_{j+1})\\ &=\sum_{j=1}^{n+1} ja_{j} (b_{n-j+1} - (n-j+2)a_{n-j+2})\\ \end{array} $
I'll leave it at this.