Let us first count how many unit squares have at last one point in common with a with an edge of length $a$: Exactly $2(a+2)$ squares. Therefore, the four edges of the chosen quadrilateral have common points with $4(a+b+4)$ unit squares. However, we double-counted some: If $a,b\geqslant 2$, we double-counted the four squares around each of the chosen vertices - resulting in the equation $4(a+b+4)-16=24\Leftrightarrow a+b=6$ with solutions $\{(3,3), (4,2)\}$ and the corresponding areas of $9$ and $8$ unit squares, respectively. If one of the sides equals $1$, w.l.o.g. $a=1$, the unit squares having points in common with the edges of the chosen rectangle form another rectangle with sidelengths $3$ and $b+2$, therefore $3(b+2)=24\Leftrightarrow b=6$. The corresponding area is $6$. So altogether, the possible areas are $\{6,8,9\}$.