Let $a, b, x$ be positive intergers such that $a \ge b$ and $a > x$.
Is it possible to express $$\left \lfloor \frac{a-b}{x} \right \rfloor $$ in terms of $\left \lfloor \frac{a}{x} \right \rfloor$ or a range to which it belongs?
If $r$ is the remainder in the divison of $a$ by $x$, we have $\frac{a-r}{x} \in \mathbb{Z}$ and $ \frac{a-r}{x} = \left \lfloor \frac{a}{x} \right \rfloor$ so $$\left \lfloor \frac{a-b}{x} \right \rfloor = \left \lfloor \frac{a-r + r - b}{x} \right \rfloor = \frac{a-r}{x} + \left \lfloor \frac{r-b}{x} \right \rfloor = \left \lfloor \frac{a}{x} \right \rfloor + \left \lfloor \frac{r-b}{x} \right \rfloor .$$
I believe that my question is equivalent to determine all possible values of $\left \lfloor \frac{r-b}{x} \right \rfloor$, but I do not know if that is possible..
Given your conditions, $\lfloor \frac{a-b}{x}\rfloor$ can be any value between $0$ (when $b=a$) and $\lfloor\frac ax\rfloor$ (when $b=0$), and that's about all you can say with the restrictions you gave.