If $S$ is a finitely generated graded algebra over $S_0$, $S_{(f)}$ is finitely generated algebra over $S_0$?

Let $\deg f=d>0$. Then $$S_{(f)}=\{\frac{x}{f^m}:\deg x=md, m\ge 0\}=S_0\oplus\frac{1}{f}S_d\oplus\frac{1}{f^2}S_{2d}\oplus\cdots.$$ But one knows that $S$ finitely generated over $S_0$ implies $S^{(d)}=S_0\oplus S_d\oplus S_{2d}\oplus\cdots$ finitely generated over $S_0$. Now the system of generators for $S_{(f)}$ over $S_0$ should be clear.

This is not an answer but it's too long for a comment. I've just came up with the following proof of the assertion that

$S^{(d)}$ is finitely generated over $S_0$.

Suppose $S$ is generated by homogeneous elements $x_i$ of degree $k_i (1 ≦ i ≦ r)$ over $S_0$. Let $n > 0$ be an integer. Let $e_i ≧ 0 (1 ≦ i ≦ r)$ be integers such that $\sum_i k_ie_i = dn$. Then the degree of $x_1^{e_1}\cdots x_r^{e_r}$ is $dn$. Suppose $e_i > d$. Then $x_1^{e_1}\cdots x_r^{e_r} = x_i^d x_1^{e_1}\cdots x_i^{e_i - d}\cdots x_r^{e_r}$. Hence $(S^{(d)})_+ = \bigoplus_{n>0} (S^{(d)})_n$ is generated as a graded $S^{(d)}$-module by $x_1^{e_1}\cdots x_r^{e_r} (0 \le e_i < d)$.
Let $U = \{x_1^{e_1}\cdots x_r^{e_r}\mid 0 \le e_i < d\}$. By induction on $n$, it is easy to see that $(S^{(d)})_n$ is generated as an $S_0$-module by finite products of elements of $U$ for every integer $n > 0$. Hence $S^{(d)}$ is generated by $U$ over $S_0$.