Glorfindel has found the answer; here is a uniqueness proof.
Write $a,b$ for the column widths; $c,d$ for the heights of the first row and the remainder of the rows together. Comparing the $24$-cell at top left and the $25$-cell at bottom right we can see that $c<d$.
we have $a+b=16$, $c+d=18$, $ac=24$, $bd=120$. Using the first two to replace $b$ with $16-a$ and $d$ with $18-c$ we get $ac=24$, $(16-a)(24-c)=120$; simplifying the latter (and using the former to get rid of its $ac$ term) we find $18a+16c=192$ or $9a+8c=96$. Writing $\alpha=9a,\gamma=8c$ we now have the sum and product of $\alpha,\gamma$, so they satisfy the quadratic equation $x^2-96x+1728=0$ which we can solve either by inspection or by plugging into the quadratic formula: $x=24,72$. We can assign these to $\alpha,\gamma$ either way around. One way yields $a=8,c=3$; the other way yields $a=8/3,c=9$. The second solution is inconsistent with the relation $c<d$ we found before, so the first must be correct.
the rest is routine.