The given relation yields $|f(x+1) - f(x)| \le 1$, so $f(x+1)-f(x) \in \{-1, 0, 1\}$. Because $f$ is one-to-one, we get $f(x+1)-f(x) \in \{-1, 1\}$, for all $x$ in $\mathbb{Z}$. We suppose WLOG that $f(1)-f(0) = 1$ (if $f$ verifies the hypothesis then $-f$ verifies it also). We have $f(2)-f(1) = \pm 1$. If $f(2)-f(1) = -1$, then $f(2) = f(0)$, a contradiction, so $f(2) = f(0) + 2$.
By easy induction it follows that $f(n) = f(0) + n$, for any positive integer $n$. Analogously we get $f(-n) = f(0) - n$, for any positive integer $n$.

Thus the functions that verify the given condition are of the form $f(x) = \pm x + k$, for any integer $x$, where $k$ is an arbitrary integer. One easily verifies that all such functions are solutions to the problem.