Let $x = 1, 2$, and we can get $f(1) = f(1) + f(2) - 1$ and $f(2) = 2f(1) + \frac{f(2)}{2} - 1$, respectively. The solution is $f(2) = 1$, $f(1) = \frac{3}{4}$.

Thus, for $x \neq 0$, there is
$$f(x) = \frac{3}{4}x + \frac{1}{x} - 1.$$
When $x \in (0, +\infty)$, $f(x) \ge 2\sqrt{\frac{3}{4}x \cdot \frac{1}{x}} - 1 = \sqrt{3} - 1$. The equal sign holds when $x = 3$.

Therefore, the minimum of $f(x)$ on $(0, +\infty)$ is $\sqrt{3} - 1$.