Simple linear recursion $x_n=\frac{x_{n-1}}{a}+\frac{b}{a}$ with $a>1, b>0$ and $x_0>0$
I tried to solve it using the generating function but it does not work because of $\frac{b}{a}$, so may you have an idea.

Using generating functions here is easy. Write your recurrence as:
$\begin{equation*}
a x_{n + 1}
  = x_n + b
\end{equation*}$
Define the generating function $g(z) = \sum_{n \ge 0} x_n z^n$, multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} x_{n + 1} z^n
  &= \sum_{n \ge 0} x_n z^n + \sum_{n \ge 0} b z^n \\
\frac{g(z) - x_0}{z}
  &= g(z) + \frac{b}{1 - z}
\end{align*}$
Solve for $g(z)$, write as partial fractions if needed:
$\begin{equation*}
g(z)
  = \frac{x_0 - (x_0 - n) z}{(1 - z)^2}
\end{equation*}$
Now you want the coefficient of $z^n$ of the above:
$\begin{align*}
[z^n] g(z)
  &= [z^n] \frac{x_0 - (x_0 - n) z}{(1 - z)^2} \\
  &= x_0 [z^n] (1 - z)^{-2} - (x_0 - b) [z^{n - 1}] (1 - z)^{-2} \\
  &= x_0 (-1)^n \binom{-2}{n} - (x_0 - b) (-1)^{n - 1} \binom{-2}{n - 1} \\
  &= x_0 \binom{n + 2 - 1}{2 - 1} - (x_0 - b) \binom{n - 1 + 2 - 1}{2 - 1} \\
  &= 2 x_0 - b + b n
\end{align*}$
