Solution:
Let $f$ be a solution to the FE. By plugging $y=\frac{x}{f(x)}$, we obtain
$$x+f(x+1)=x f\left(x+\frac{x}{f(x)}\right)+x \Longleftrightarrow f(x+1)=x f\left(x+\frac{x}{f(x)}\right)$$
Plug in $x=1$ in (1) to get
$$f(2)=f\left(1+\frac{1}{f(1)}\right)$$
Note that if $f$ is injective, one can easily finish as $f(1)=1$ by the previous equality. By plugging in $x=1$ in the original equation, we get
$$1=y f(y) \Longleftrightarrow f(y)=\frac{1}{y}, \quad \forall y \in \mathbb{R}_{>0}$$
which is indeed a solution to the equation:
$$x+f\left(\frac{y}{x}+1\right)=\frac{x}{x+y}+y f\left(\frac{y}{x}\right) \Longleftrightarrow x+\frac{x}{x+y}=\frac{x}{x+y}+x, \quad \forall x, y \in \mathbb{R}_{>0}$$
Now, it remains to prove that $f$ is injective: Assume that there exist $u, v \in \mathbb{R}_{>0}$, with $u>v$ such that $f(u)=f(v)$. Rewriting the initial equation, we get:
$$x \cdot(1-f(x+y))=y f(y f(x))-f(y f(x)+1) .$$
Substituting $x=u$ and $x=v$ in the latter equation, we get that
$$u \cdot(f(u+y)-1)=v \cdot(f(v+y)-1), \quad \forall y \in \mathbb{R}_{>0}$$
and by replacing $y \rightarrow y-v$ and introducing $C=u-v>0$, we get
$$u \cdot(f(C+y)-1)=v \cdot(f(y)-1), \quad \forall y>v$$
Let $y_{0}$ be a fixed constant such that $v<y_{0} \leq v+C$ and let's analyse the sequence $\left(f\left(y_{0}+n C\right)\right)_{n \in \mathbb{N}}$. If $f\left(y_{0}\right)=1$, we get that the sequence $\left(f\left(y_{0}+n C\right)\right)_{n \in \mathbb{N}}$ is constant and every term is equal to 1.
Now, if $f\left(y_{0}\right) \neq 1$, by (5), observe that
$$\frac{u}{v}=\frac{f\left(y_{0}+n C\right)-1}{f\left(y_{0}+(n+1) C\right)-1} \quad \forall n \in \mathbb{N}$$
Therefore, by induction or telescopic product, we get
$$f\left(y_{0}+n C\right)-1=\left(\frac{v}{u}\right)^{n} \cdot\left(f\left(y_{0}\right)-1\right)$$
Hence, by making $n \rightarrow \infty$ and using $v<u$, we get that the sequence $\left(f\left(y_{0}+n C\right)\right)_{n \in \mathbb{N}}$ converges to 1. From these two cases, we conclude that the sequence $\left(f\left(y_{0}+n C\right)\right)_{n \in \mathbb{N}}$ converges to 1 for any $v<y_{0}$. Now, let $a_{n}=y_{0}+n C$ and let's make the substitution $y=\frac{a_{n}}{f(x)}$. We get
$$x+f\left(a_{n}+1\right)=x f\left(x+\frac{a_{n}}{f(x)}\right)+\frac{a_{n}}{f(x)} \cdot f\left(a_{n}\right)$$
and let's fix $x=c$. Then, when
$$n \rightarrow+\infty, \quad c+f\left(a_{n}+1\right) \rightarrow c+1$$
because the sequence
$$\left(f\left(\left(y_{0}+1\right)+n C\right)\right)_{n \in \mathbb{N}}=\left(f\left(a_{n}+1\right)\right)_{n \in \mathbb{N}}$$
converges to 1 . However, we have
$$c f\left(c+\frac{a_{n}}{f(c)}\right)+\frac{a_{n}}{f(c)} \cdot f\left(a_{n}\right) \rightarrow+\infty$$
because
$$\frac{a_{n}}{f(c)} \cdot f\left(a_{n}\right) \rightarrow+\infty \text{ and } c f\left(c+\frac{a_{n}}{f(c)}\right)>0$$
Thus, we get a contradiction by (7).