You cannot recover a convex polytope from its projection areas, even if you know the whole function (unit vector) $\mapsto$ (area of projection along this vector). There exist two different polytopes $P_1$ and $P_2$ such that for every unit vector $v\in\mathbb R^3$, the areas of the two projections along $v$ are equal.

Let me begin with a two-dimensional example. In this case, the projections are one-dimensional, and the "projection area" is width. Consider a regular triangle $T$ with side 1 and regular hexagon $H$ with side 1/2, positioned so that their sides are parallel. Their widths are the same in every direction. One can prove this without computation: the hexagon is the Minkowski symmetrization of the triangle and the symmetrization preserves widths.

Now go to dimension 3. Let $P_1=T\times[0,1]$, the prism with height 1 based on the regular triangle with side 1. Let $P_2=(\sqrt{2/3}H)\times[0,\sqrt{3/2}]$, the prism with height $\sqrt{3/2}$ based on the regular hexagon with side $1/\sqrt6$ (I hope that I get the constants right). I claim that they have the same projection area in every direction.

More generally, consider a prism of height $h$ based on a convex figure $F$ of area $A$ such that the base is parallel to the $xy$-plane. Its area of projection along a vector $v=(\cos\alpha\cos\theta,\sin\alpha\cos\theta,\sin\theta)$, is given by $$ A\cdot \sin\theta+w(\alpha)\cdot h \cdot \cos\theta $$ where $w(\alpha)$ is the width of $F$ in the horizontal direction that forms oriented angle $\alpha$ with the $x$-axis. The constants in the above example are chosen so that the two prisms have the same $A$ and the difference in $w(\alpha)$ is compensated by the difference in $h$.