Firstly, we will prove that $\frac{f(x)}{x}$ is a constant.
Assume that there exists $a, b \in (0, +\infty)$ such that $\frac{f(a)}{a} \neq \frac{f(b)}{b}$.
Without loss of generality, we assume that $\frac{f(a)}{a} < \frac{f(b)}{b}$. By plugging $x = a, x = b$ into the relation, we get
$$f\left(y + \frac{f(a)}{a}\right) = f\left(y + \frac{f(b)}{b}\right) = f(y) + 1$$
for all positive real numbers $y$. Denote $K = \frac{f(b)}{b} - \frac{f(a)}{a}$, we obtain that $f(x) = f(x + K)$ for all $x > \frac{f(a)}{a}$. Furthermore, we have
$$f\left(y + n \cdot \frac{f(a)}{a}\right) = f\left(y + (n-1) \cdot \frac{f(a)}{a}\right) + 1 \\ = \dots = f(y) + n > n.$$
Thus $f(x) > n$ for all $x > n \cdot \frac{f(a)}{a}$. Now, for an arbitrary positive number $x$, choose $n = \lfloor f(x) \rfloor + 2 > f(x)$ and a positive integer $m$ such that $x + mK > n \cdot \frac{f(a)}{a}$, we have $f(x + mK) > n > f(x) = f(x + mK)$, which is a contradiction.

Hence, $\frac{f(x)}{x}$ is a constant. Denote $\frac{f(x)}{x} = c$ and replace $f(x) = cx$ in the original equation, one can find that $c = 1$. Thus, $f(x) = x$ for all positive numbers $x$. $\square$