, 2018

Solution:
Let $P(x, y)$ denote the assertion that
$$f(x+f(y+x y))=(y+1) f(x+1)-1.$$

Claim 1. $f$ is injective.

Proof: If $f(a)=f(b)$ then $P\left(x, \frac{a}{x+1}\right), P\left(x, \frac{b}{x+1}\right)$ yields $a=b$, since $f(x+1) \in \mathbb{R}^{+}$ so in particular is nonzero.

Now $P\left(x, \frac{1}{f(x+1)}\right)$ yields
$$f\left(x+f\left(\frac{x+1}{f(x+1)}\right)\right)=f(x+1)$$
hence by injectivity
$$x+f\left(\frac{x+1}{f(x+1)}\right)=x+1$$
so that
$$f\left(\frac{x+1}{f(x+1)}\right)=1$$
By injectivity, this equals some constant $c$, so that
$$\frac{x+1}{f(x+1)}=c$$
for all $x \in \mathbb{R}^{+}$.

Now letting $x, y>1$ in $P(x, y)$ automatically yields
$$\frac{x}{c}+\frac{y+x y}{c^{2}}=(y+1)\left(\frac{x+1}{c}\right)-1$$
which immediately yields $c=1$ if we take $x, y$ large.

Finally, we have $f(x+1)=x+1$ for all $x \in \mathbb{R}^{+}$.

Finally, $P\left(x, \frac{y}{x+1}\right)$ yields
$$f(x+f(y))=\left(\frac{y}{x+1}+1\right) f(x+1)-1=x+y$$
so that fixing $y$ and letting $x>1$ yields
$$x+f(y)=x+y$$
so that $f(y)=y$.

This was for arbitrary positive $y$, so that $f(x)=x$ for all $x \in \mathbb{R}^{+}$, which clearly works.