Let's restart from beginning.
We have
$$
\left\{ \matrix{
a_{\,n} = 0\quad \left| {\;n < 0} \right. \hfill \cr
a_{\,0} = 1 \hfill \cr
a_{\,1} = 3 \hfill \cr
a_{\,n} = 2a_{\,n - 1} - 3a_{\,n - 2} + 4n - 1 \hfill \cr} \right.
$$
The best approach is to adjust the recurrence as to accomodate
the initial values. We shall do in this way
$$
\eqalign{
& a_{\,0} = 1 = 2a_{\,0 - 1} - 3a_{\,0 - 2} + 4 \cdot 0 - 1 = - 1\quad \Rightarrow \cr
& \Rightarrow \quad a_{\,n} = 2a_{\,n - 1} - 3a_{\,n - 2} + 4n - 1 + 2\left[ {0 = n} \right] \cr
& a_{\,1} = 3 = 2a_{\,0} - 3a_{\,1 - 2} + 4 \cdot 1 - 1 + 2\left[ {0 = 1} \right] = 5\quad \Rightarrow \cr
& a_{\,n} = 2a_{\,n - 1} - 3a_{\,n - 2} + 4n - 1 + 2\left[ {0 = n} \right] - 2\left[ {1 = n} \right] \cr}
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
So we get a recurrence which is valid for all $0 \le n$.
$$
\left\{ \matrix{
a_{\,n} = 0\quad \left| {\;n < 0} \right. \hfill \cr
a_{\,n} = 2a_{\,n - 1} - 3a_{\,n - 2} + 4n - 1 + 2\left[ {0 = n} \right] - 2\left[ {1 = n} \right] \hfill \cr} \right.
$$
or better
$$
a_{\,n} = 2\left[ {1 \le n} \right]a_{\,n - 1} - 3\left[ {2 \le n} \right]a_{\,n - 2} + 4n - 1 + 2\left[ {0 = n} \right]
- 2\left[ {1 = n} \right]\quad \left| {\;0 \le n} \right.
$$
Now we can multiply by $x^n$ and sum
$$
\eqalign{
& f(x) = \sum\limits_{n = 0}^\infty {a_{\,n} x^{\,n} } = \cr
& = 2\sum\limits_{n = 0}^\infty {\left[ {1 \le n} \right]a_{\,n - 1} x^{\,n} } - 3\sum\limits_{n = 0}^\infty {\left[ {2 \le n} \right]a_{\,n - 2} x^{\,n} } + \cr
& + 4\sum\limits_{n = 0}^\infty {nx^{\,n} } - \sum\limits_{n = 0}^\infty {1x^{\,n} } + 2\sum\limits_{n = 0}^\infty {\left[ {0 = n} \right]x^{\,n} } - 2\sum\limits_{n = 0}^\infty {\left[ {1 = n} \right]x^{\,n} } = \cr
& = 2\sum\limits_{n = 1}^\infty {a_{\,n - 1} x^{\,n} } - 3\sum\limits_{n = 2}^\infty {a_{\,n - 2} x^{\,n} } + \cr
& + 4\sum\limits_{n = 0}^\infty {nx^{\,n} } - \sum\limits_{n = 0}^\infty {1x^{\,n} } + 2x^{\,0} - 2x^{\,1} = \cr
& = 2\sum\limits_{n = 0}^\infty {a_{\,n} x^{\,n + 1} } - 3\sum\limits_{n = 0}^\infty {a_{\,n} x^{\,n + 2} }
+ 4\sum\limits_{n = 0}^\infty {nx^{\,n} } - \sum\limits_{n = 0}^\infty {x^{\,n} } + 2 - 2x = \cr
& = \left( {2x - 3x^{\,2} } \right)\sum\limits_{n = 0}^\infty {a_{\,n} x^{\,n} } + 4\sum\limits_{n = 0}^\infty {nx^{\,n} } - {1 \over {1 - x}} + 2 - 2x \cr}
$$
And the term with $nx^n$ becomes
$$
\eqalign{
& \sum\limits_{n = 0}^\infty {nx^{\,n} } = \sum\limits_{n = 1}^\infty {nx^{\,n} } = x\sum\limits_{n = 1}^\infty {nx^{\,n - 1} } = \cr
& = x{d \over {dx}}\sum\limits_{n = 0}^\infty {x^{\,n} } = {x \over {\left( {1 - x} \right)^{\,2} }} \cr}
$$
Finally
$$
\eqalign{
& f(x) = \left( {2x - 3x^{\,2} } \right)f(x) + 4{x \over {\left( {1 - x} \right)^{\,2} }} - {1 \over {1 - x}} + 2 - 2x \cr
& f(x) = {{5x - 1} \over {\left( {1 - 2x + 3x^{\,2} } \right)\left( {1 - x} \right)^{\,2} }} + {{\left( {1 - x} \right)} \over {\left( {1 - 2x + 3x^{\,2} } \right)}} \cr}
$$