Plugging $y = -x$ and $d = a - b$ yields $\lfloor d x \rfloor + \lfloor -d x \rfloor = 0$, for every $x \in \mathbb{R}$. (*)
If $d \neq 0$, then (*) is false for $x = 1/(2d)$, hence $d = 0$. Then $x + y = 1$ leads to $2\lfloor a \rfloor = 2a$, therefore $a$ is an integer.
If $a = 0$ the relation is fulfilled, therefore a solution is $a = b = 0$.

If $a \neq 0$, relation $2a\lfloor x + y \rfloor = 2\lfloor a(x + y) \rfloor \le 2a(x + y)$ with $x + y = 1/2$ leads to $a \ge 0$, therefore $a \ge 1$. Now $x + y = 1/a$ gives $1 = a \lfloor 1/a \rfloor$, whence $a = b = 1$, which is the second solution.