$P(0,x): f(1)=x+f(f(0)f(y))$-Injective $P(x,0): f(xf(0)+1)=f(f(x)f(0))$ $xf(0)+1=f(x)f(0)$ $f(0)=0 \rightarrow 1=0.$ Contraction. $f(0)\neq 0 \rightarrow f(x)=x+c$ And after checking we get the only solution $f(x)=x-1$.

Let $P(x,y)$ be the assertion of $f(xf(y)+1)=y+f(f(x)f(y))$. $P(0,0)\implies f(1)=f(f(0)^2)$ Fix $x$ and let $f(a)=f(b)$, then $P(x,a)$ and $P(x,b)$ give us that $a=b$, injectivity. Thus, $f(0)^2=1\implies f(0)=-1\text{ or } f(0)=1$. If $f(0)=-1$, then $P(x,0)\implies f(-x+1)=f(-f(x))\implies f(x)=x-1$. If $f(0)=1$, then $P(x,0)\implies f(x+1)=f(f(x))\implies f(x)=x+1$. If $f(x)=x-1$, then $\mathcal{LHS}=f(xf(y)+1)=f(x(y-1)+1)=f(xy-x+1)=xy-x$ and $\mathcal{RHS}=y+f(f(x)f(y))=y+f(xy-x-y+1)=y+xy-x-y=xy-x$, thus $\mathcal{LHS}=\mathcal{RHS}$. If $f(x)=x+1$, then $\mathcal{LHS}=f(xf(y)+1)=f(x(y+1)+1)=f(xy+x+1)=xy+x+2$ and $\mathcal{RHS}=y+f(f(x)f(y))=y+f(xy+x+y+1)=y+xy+x+y=xy+x+2y+2$, thus $\mathcal{LHS}\neq\mathcal{RHS}$. Answer. $\boxed{f(x)=x-1\,\forall x\in \mathbb R}$.

Denote the assertion by $P(x,y).$ $P(0,x)\implies f(1)=x+f(f(x)f(0))$ this means bijective. $P(x,0)\implies f(x)f(0)=xf(0)+1\iff f(x)=x+c.$ Checking gives $f(x)\equiv x-1$ only.

The only solution is $f(x) = x - 1$, which works. Now we prove it's the only one. Let $P(x,y)$ denote the given assertion. If $f(a) = f(b)$, then $P(x,a)$ compared with $P(x,b)$ gives that $a = b$, so $f$ is injective. $P(x,0): f(xf(0) + 1) = f(f(x)f(0))$, so $f(x)f(0) = xf(0) + 1$. If $f(0) = 0$, then we have a contradiction, so $f(0) \ne 0$, so dividing both sides by $f(0)$ gives $f(x) = x + \frac{1}{f(0)}$. Now, setting $x = 0$ gives $f(0) = \frac{1}{f(0)} $, so $f(0) \in \{-1,1\}$. $P(0,x): f(1) = x + f(f(0) f(x))$. If $f(0) = 1$, then $f(x) = x + 1$, so $f(1) = x + f(f(x)) = 2x + 2$, so $x = 0$, absurd. Hence $f(0) = -1$, and $f(x) = x -1$.