Well, the other two regions are shielded from the given charge by perfect conductors. These shield the electric field and it is zero beyond them.
A bit more rigorously, say you're in the second quadrant ($x<0$, $y>0$). Of course, the image charge at $(-a,b)$ is not present, but it doesn't mean its effect isn't. The image charge is actually a nice representation for the surface charge induced by your original charge at $(a,b)$ on the vertical conducting plane. This surface charge looks like (is indistinguishable from) a point charge a length $a$ behind the conducting plane, so in the first quadrant it looks like a negative charge at $(-a,b)$, but on the second quadrant it sits on top of, and cancels, the original charge.
A similar thing happens (must happen) with the image charges in the third and fourth quadrant. The one at $(a,-b)$ is a representation of a surface charge centered around $(a,0)$, and therefore looks like the same image charge at $(a,-b)$ as seen from the second quadrant. This is cancelled in the second quadrant by a surface charge around the origin, on the positive $x$ and $y$ axes, which is represented (from the first quadrant) by the third-quadrant image charge. Since the total field must cancel, this third induced charge must look like (from the second quadrant) as a positive charge sitting at $(a,-b)$, though I don't know of a simple explanation for this.