When $\lfloor{ab}\rfloor = \lfloor{a}\rfloor\lfloor{b}\rfloor$

As suggested in the comments, let's expand the LHS a little. I'll use $\langle a\rangle$ to represent the fractional part of $a$. $$\lfloor{ab}\rfloor = \lfloor{(\lfloor{a}\rfloor + \langle a\rangle)(\lfloor{b}\rfloor + \langle b\rangle)}\rfloor = \lfloor{a}\rfloor \lfloor{b}\rfloor + \big\lfloor \lfloor{a}\rfloor \langle b\rangle + \lfloor{b}\rfloor \langle a\rangle + \langle a \rangle \langle b\rangle \big\rfloor.$$

Thus a necessary and sufficient condition for the equation you propose to hold is that $$0 < \lfloor{a}\rfloor \langle b\rangle + \lfloor{b}\rfloor \langle a\rangle + \langle a \rangle \langle b\rangle < 1.$$

In the first quadrant, in the square tile $[A,A+1)\times[B,B+1)$, we have

$$\lfloor a\rfloor \lfloor b\rfloor=AB$$ and $$AB\le ab<(A+1)(B+1).$$

We are looking for the points such that

$$AB\le ab<AB+1,$$ which form a subset of the tile delimited by the equilateral hyperbola $$ab=AB+1.$$

For all tiles such that $AB=0$, the solutions cover the area $$0\le ab<1,$$ i.e. the space between the axis and $ab=1$.

For the other tiles, the intersections of the hyperbola with the edges are

$$\begin{align} a&=A\to &Ab=AB+1\to &b=B+\frac1A,\\ a&=A+1\to&(A+1)b=AB+1\to &b=\frac{AB+1}{A+1}\le B,\\ b&=B\to &aB=AB+1\to &b=A+\frac1B,\\ b&=B+1\to &a(B+1)=AB+1\to &b=\frac{AB+1}{B+1}\le A.\\ \end{align}$$

So the solution area is the inside of the curvilinear triangle with straight sides $a=A,b=B$ and the branch of hyperbola $ab=AB+1$, from $(A,B+1/A)$ to $(A+1/B,B)$. The area of the triangle is roughly proportional to $1/AB$.

The study will be similar for the other quadrants.