This theorem is not true if $b$ is not an integer. Take $x=b=1.5$ and take $a=1$.
If $b$ is an integer, this follows from the rule $$\left\lfloor \frac{y}{b}\right\rfloor = \left\lfloor \frac{\lfloor y \rfloor}b\right\rfloor$$
Setting $y=\frac x a$.
Showing this rule, then, suffices.
Let $y = \lfloor y \rfloor + \{y\}$, where $0\leq \{y\} < 1$.
Use division algorithm to write $\lfloor y \rfloor = qb + r$ with $0\leq r
The $\frac{\lfloor y \rfloor} b = q + \frac{r}{b}$, and $0\leq \frac{r}{b} <1$, so
$$\left\lfloor \frac{\lfloor y \rfloor}b\right\rfloor = q$$
On the other hand, since $[y]< (q+1)b$, since $0\leq\{y\}<1$, then $y = [y]+\{y\}<(q+1)b$.
So $q \leq \frac y b < q+1$ and again $$\left\lfloor \frac{y}{b}\right\rfloor = q $$