The function $f(x)=\frac{\lfloor x\rfloor -\{x\}}{2}$ is surjective and works. Here $x=\lfloor x \rfloor +\{x\}.$

The function $f(x)= \begin{cases} x &x<0\\x-1 &x\geq 0 \\\end{cases}$ is clearly surjective. It also works by considering all cases: Case 1. $x<0$, $y<0:$ $f(x+y)-f(x)-f(y)=(x+y)-x-y=0$ Case 2. $x<0$, $y\geq 0$ (or $x\geq 0$, $y<0$)$:$ $f(x+y)-f(x)-f(y)=(x+y+\varepsilon)-x-(y-1)=\varepsilon+1\in\{0,1\}$ where $\varepsilon\in\{-1,0\}$. Case 3. $x\geq 0$, $y\geq 0:$ $f(x+y)-f(x)-f(y)=(x+y-1)-(x-1)-(y-1)=1$ Therefore $f(x+y)-f(x)-f(y)$ only takes on values $0$ and $1$ and both values appear.