Here is my progress: Let $P(x, y)$ be the given proposition and $c = f(1)$. $P(1, \frac{1}{c})$ implies that $f(2) = f(\frac{1}{c}+1)$. Injectivity would imply $c = 1$ and from $P(1, x)$, we would conclude that $f(x) = \frac{1}{x}$ for all $x$. I haven't been able to prove that $f$ is injective or find another way to show that $f(1) = 1$ yet.

Wikiliks wrote: Here is my progress: Let $P(x, y)$ be the given proposition and $c = f(1)$. $P(1, \frac{1}{c})$ implies that $f(2) = f(\frac{1}{c}+1)$. Injectivity would imply $c = 1$ and from $P(1, x)$, we would conclude that $f(x) = \frac{1}{x}$ for all $x$. I haven't been able to prove that $f$ is injective or find another way to show that $f(1) = 1$ yet. I'll just show here that $f$ is injective, and then one can use the method described above. Suppose $f$ is not injective. Then there exists $x_0, x_1$ two positive reals such that $x_0 < x_1$ and $f(x_0) = f(x_1)$. For all $y>0$, rewriting $P(x_0,y)$ and $P(x_1,y)$ gives: $$x_0f(x_0+y) - x_0 = f(yf(x_0) +1) - yf(yf(x_0))=f(yf(x_1) +1) - yf(yf(x_1))=x_1f(x_1+y) - x_1 $$This means that $f(y+x_1) = \frac{x_0}{x_1}f(y+x_0) + 1 - \frac{x_0}{x_1}$ for all $y >0$, and if $d=x_1-x_0$ can be rewritten as: $f(y+d) = \frac{x_0}{x_1}f(y) + 1 - \frac{x_0}{x_1}$ for all $y>x_0$. This implies 1) $f(y)> 1 - \frac{x_0}{x_1} > 0$ for all $y>x_1$ and 2) for all $y>0$, the sequence $f(y+nd)$ converges (towards $1$) as $n$ goes to infinity Let then $x, y$ be two positive reals such that $yf(x)>x_1$, and $n$ any positive integer. $P(x, y+n\frac{d}{f(x)})$ gives: $$x+f(yf(x) + 1 +nd) = xf(x+y+n\frac{d}{f(x)}) +(y+n\frac{d}{f(x)}))f(yf(x) + nd) > nd \frac{f(yf(x)+nd)}{f(x)} >\frac{nd}{f(x)}(1-\frac{x_0}{x_1})$$And according to 2) above, the LHS $x+f(yf(x)+1+nd)$ converges towards $x+1$ as $n$ goes to infinity, but the RHS $\frac{nd}{f(x)}(1-\frac{x_0}{x_1})$ diverges towards infinity, so we have a contradiction, meaning $f$ is indeed injective.

My complete solution, also another proof for injective (i think so). Claim 1: $f$ is not bounded above. Proof. From $P\left(x,\frac{y}{f(x)} - 1\right)$ it follows that \[f(y) = xf\left(x + \frac{y-1}{f(x)}\right) + \frac{y - 1}{f(x)}f(y - 1) - x,\forall x > 0, y > 1 \tag{2} \]From $(2)$ as $y \to +\infty$, the right-hand side $\to +\infty$ so $\displaystyle \lim_{y \to +\infty} f(y) = +\infty$. Claim 2: $f$ is injective on $\mathbb{R^+}$ Proof. Suppose $f(a)=f(b)$ and $a > b$. Let $d = a - b, q = \frac{b}{a}, r= \frac{d}{a}$, we have $d,q,r > 0$ and $q < 1$. Substituting $P(a,x), P(b,x) \rightarrow a - af(a + x) = b - bf(b + x)$. As $x \to x -b$ and let $\delta = 4b$ be sufficiently large, we get \[ f(x + d) = qf(x) + r, \forall x > \delta \tag{$\clubsuit$ } \] We state and prove the following lemma: Lemma. Consider the function $f: \mathbb{R^+} \to \mathbb{R^+}$ satisfying \[ f(x + d) = qf(x) + r, \forall x > M \]with $M$ being sufficiently large positive real number with $q < 1$ and $d, r > 0$. Then $\displaystyle \lim_{x \to +\infty} f(x) = \frac{r}{1 - q}$. Proof. Substituting $x \to x + d$, we get \[f(x + 2d) = qf(x + d) + r = q(qf(x) + r) + r = q^2f(x) + qr + r, \forall x > \delta\]By induction it's easy to prove that \[f(x + nd) = q^nf(x) + r\sum_{i = 0}^n q^i, \forall x > \delta \tag{3}\]From $(3)$ we rewrite \[ f(x + nd) = q^nf(x) + r.\frac{1 - q^{n}}{1 - q}, \forall x > \delta \tag{4} \]From $(4)$, as $n \to +\infty$ with $q < 1$, we have \[ \lim_{x \to +\infty} f(x) = \displaystyle \lim f(x + nd) = \frac{r}{1 - q} \]This completes the proof. Applying the above lemma to $(\clubsuit)$, we get $\displaystyle \lim_{x \to +\infty} f(x) = \frac{r}{1 - q}$. But according to Claim 2, $f$ is not bounded above and $\displaystyle \lim_{x \to +\infty} f(x) = +\infty$, contradiction. Therefore $q = 1$ or $d = 0 \rightarrow a = b$. Hence $f$ is injective on $\mathbb{R^+}$. Setting $P(1,1)$ we get $f(2) = f(1 + \frac{1}{f(1)})$. Since $f$ is injective, $f(1) = 1$. Give $P(1,x)$ the we get the satisfied function is $\boxed{f(x) = \frac{1}{x}, \forall x > 0}$.