, 2020

Denote $xy = z$, then the variables $y$ and $z$ can take all positive real values independently. For these variables the equality has the form
$$\frac{z}{y}g\left(\frac{z}{y} + g(y)\right) = g(g(z) + 1). \qquad (1)$$
Fix any value of $z$ and vary $y$ over $\mathbb{R}_+$, then $g(g(z) + 1)$ is fixed while $\frac{z}{y}$ attains all positive real values, hence $g(\frac{z}{y} + g(y))$ attains all positive real values, i.e. $g$ is surjective.
Suppose $\frac{g(y_2) - g(y_1)}{y_2 - y_1} > 0$ for some positive reals $y_2 \neq y_1$. Substitute $z = y_2 y_1 \cdot \frac{g(y_2) - g(y_1)}{y_2 - y_1}$ into (1) firstly with $y = y_2$, secondly with $y = y_1$. In both cases $g(\frac{z}{y} + g(y)) = g(\frac{y_2 g(y_2) - y_1 g(y_1)}{y_2 - y_1})$, hence we will obtain that $y_1 = y_2$ – a contradiction. Thus $\frac{g(y_2) - g(y_1)}{y_2 - y_1} \le 0$ for all positive reals $y_2 \neq y_1$ which means that $g$ is non-strictly decreasing.
Since $g$ is surjective and non-strictly decreasing, it's continuous and, in particular $\lim_{y \to +\infty} g(y) = 0$. Now consider the initial equation
$$xg(x + g(y)) = g(g(xy) + 1),$$
when $y \to +\infty$. Then $g(x + g(y)) \to g(x)$ and $g(xy) + 1 \to 1$. Hence $xg(x) = g(1)$ and $g(x) = \frac{c}{x}$, where $c = g(1)$. Straightforward verification shows that such function satisfy the problem conditions for any $c \in \mathbb{R}$.