First, replace $x \to \frac{z}{y}$ then
$$f(f(z) + 1) = \frac{z}{y} f\left(\frac{z}{y} + f(y)\right), \forall y, z > 0. \quad (1)$$
In (1), continue to set $z = 1$ and set $f(f(1) + 1) = c$, we have $f\left(\frac{1}{y} + f(y)\right) = c y$ for all $y > 0$. The right-hand side takes on any values on $\mathbb{R}^+$, so the function $f(y)$ is surjective on $\mathbb{R}^+$.

Now suppose that there exists $0 < y_1 < y_2$ such that $f(y_1) < f(y_2)$. We choose $z_0$ such that
$$\frac{z_0}{y_1} + f(y_1) = \frac{z_0}{y_2} + f(y_2) \iff z_0 = y_1 y_2 \frac{f(y_2) - f(y_1)}{y_2 - y_1} > 0.$$
In (1), replace $z = z_0$ and let $y = y_1, y = y_2$ respectively, we have
$$f(f(z_0) + 1) = \frac{z_0}{y_1} f\left(\frac{z_0}{y_1} + f(y_1)\right) = \frac{z_0}{y_2} f\left(\frac{z_0}{y_2} + f(y_2)\right).$$
Base on the choice of $z_0$, we have
$$f\left(\frac{z_0}{y_1} + f(y_1)\right) = f\left(\frac{z_0}{y_2} + f(y_2)\right),$$
it follows that $y_1 = y_2$. This contradiction shows that with $0 < y_1 < y_2$, there must be $f(y_1) \ge f(y_2)$ so $f$ is a non-increasing function. Thus, $f$ is both surjective and monotone on $\mathbb{R}^+$ so it will be continuous on $\mathbb{R}^+$. Since $f$ is continuous and non-increasing, there exists $\lim_{x \to +\infty} f(x) = c \ge 0$. However, if $c > 0$ then $f(x) \ge c > 0$, $\forall x \in \mathbb{R}^+$, which contradicts surjectiveness. Therefore $c = 0$.

Finally, in the given condition, for each $x > 0$, given $y \to +\infty$ then $f(f(xy) + 1) \to f(1)$ and $f(x + f(y)) \to f(x)$. These imply that $f(1) = x f(x)$ or $f(x) = \frac{c}{x}$ with $c = f(1) > 0$. It is easy to check that this function satisfies the problem. $\square$