We claim that the only solution is $\boxed{f(x)\equiv x}$, which readily checks. Denote the equation by $P(x,y)$. $P(0,1)\implies (f(0)+1)(f(-1)+1)=f(0)$ and $P(0,-f(0))\implies f(f(0)+f(0)f(-f(0)-1))=0$, so let $a=f(0)+f(0)f(-f(0)-1)$. Now $P(a,a+1)\implies (a+1)(f(-1)+1)=0$. It is easy to see that $a=-1$ gives a contradiction, so we have $f(-1)=-1$ and also then $f(0)=0$. $P(x,0)\implies f(x)^2+f(x)=f(f(xf(x+1)))$ and $P(x,1)\implies (f(x)+1)(f(x-1)+1)=f(x)^2+f(x)$ so $(f(x)+1)(f(x-1)-f(x)+1)=0~(\star)$. Claim 1. $f(x)=0\iff x=0$. Proof. Taking the difference of $P(-1,-x)$ and $P(0,-x)$, we get $xf(x)=(x+1)f(x-1)+1~(2)$. Take $u$ so that $f(u)=0$. Then taking $x=u$ in $(1)$, we get $f(u-1)=-1$ and taking $x=u$ in $(2)$, we get $0=-(u+1)+1\iff u=0$, as claimed. $\blacksquare$ Claim 2. $f(x)=-1\iff x=-1$. Proof. Suppose $f(u)=-1$. Now by $P(u,0)$, $$f(f(uf(u+1)))=0\iff f(uf(u+1))=0 \iff uf(u+1)=0,$$from which it's clear that $u=-1$. $\blacksquare$ Thus for $x\neq-1$, $(\star)$ becomes $f(x-1)-f(x)=-1$. We show that this holds at $x=-1$ too: Claim 3. $f(-2)=-2$. Proof. We have $f(2)=f(1)+1=f(0)+2=2$. Now $$P(-2,0)\implies f(-2)^2+f(-2)=f(f(-2f(-1)))=f(f(2))=2,$$so $f(-2)=1$ or $f(-2)=-2$. If the first case were true, $P(0,-2)\implies f(2)=-4$, contradiction. $\blacksquare$ $P(0,x)\implies x(f(-x)+1)=f(-xf(x-1))$. $P(-1,x)\implies (x-1)f(-x)=(x-1)(f(-1-x)+1)=f(-xf(x-1))$. Combining this with the line above, $$x(f(-x)+1)=(x-1)f(-x)\iff f(x)=x.$$

My solution: Let $P(x,y)$ be the assertion of our functional equation. We can find easily $P(x,-f(x))\implies \exists a,f(a)=0.$ Then from $P(a,0)\implies f(f(af(a+1))=0,$ and $P(a,a+1)\implies (a+1)(f(-1)+1)=f(f(af(a+1))=0.$ We can find $a\not = -1,$ Then $f(-1)=-1,$ and $a\not = -1,$ and $f(0)=0.$ $P(x,-1)\implies f(x)f(x-1)+f(x-1)-1=f(x)^2.$ $(1)$ Also $P(0,-x)$ and $P(-1,-x)$ gives $xf(x-1)=xf(x)-f(x-1)-1.$ $(2)$ From $(1),(2)$ we find $(f(x)-f(x-1))(f(x)-x)=0.$ $*$ Lemma: $f(x)=0\iff x=0.$ Proof: In $(2)$ give $x=a$ gives $(a+1)f(a-1)=-1 \iff f(a-1)=-1\iff a=0.$ As desired. Then from $P(x,-f(x))$ gives $f(xf(x+1))=-f(x)f(-f(x)-1)$ and from $P(x,x+1)\implies f(xf(x+1))=(x+1)f(x),$ Then $(x+1)f(x)=-f(x)f(-f(x)-1)\implies f$ is injective. Then $f(x)\not = f(x-1).$ Then from $*$ we get $f(x)=x,$ which is satisfied.

Nice problem

Let $P(x,y)$ denote the assertion $(f(x)+y)(f(x-y)+1)=f(f(xf(x+1))-yf(y-1)).$ The identity function clearly works. $P(0,-f(0))$ implies $f(x_0)=0$ for some $x_0.$ Then $P(x_0,0)$ implies $f(f(x_0f(x_0+1)))=0.$ So $P(x_0,x_0+1)$ implies $f(-1)\in \{0,-1\}.$ And $P(0,1)$ shows $f(-1)=-1$ and $f(0)=0.$ \begin{align*} P(0,-x)\implies & -x(f(x)+1)=f(xf(-x-1))\\ P(-1,-x)\implies & -(x+1)(f(x-1)+1)=f(xf(-x-1))\\ \therefore & (x+1)(f(x-1)+1)=x(f(x)+1)\quad (\dagger) \\ P(x,0)\implies & f(x)(f(x)+1)=f(f(xf(x+1))) \quad (*) \\ P(x,1)\implies & (f(x)+1)(f(x-1)+1)=f(f(xf(x+1)))\\ \therefore & f(x)(f(x)+1)=(f(x-1)+1)(f(x)+1) \quad (\heartsuit) \end{align*}For any $f(x_0)=0$, $(\heartsuit)$ and $(\dagger)$ gives $x_0=0.$ Let $f(y_0)=-1$ then, $(*)$ shows $y_0=-1.$ So for all $x\neq -1$, comparing $(\heartsuit)$ and $(\dagger)$ we are done.