First, taking $x=-1$, we find $f(f(0))=0$. So, $f(x)=0$ has at least one solution, i.e. $x=f(0)$. Next, I show $f$ is injective which will prove uniqueness. Let $x_1,x_2$ be such that $f(x_1)=f(x_2)$. Taking $x_1-1$ and $x_2-1$, we find $(x_1-1)^3 +1 = (x_2-1)^3+1$. As $t\mapsto t^3$ is injective (easily verified), this means $x_1=x_2$, proving that $f$ is injective.

A much more interesting question is whether such a function exists. And indeed, $f(x)=\mbox{sgn}(x-1)|x-1|^{\sqrt{3}} + 1$, where $\mbox{sgn}(\cdot)$ is $1$, $0$ or $(-1)$ for positive, zero and negative argument respectively, works.