Solution:

Let $\pi$ be the plane through $A$ perpendicular to the edge $AB$, and let $C'$, $D'$ be the projections of $C$ and $D$ onto $\pi$. The projection of the tetrahedron $ABCD$ onto $\pi$ is the triangle $AC'D'$.

The volume of the tetrahedron $ABCD$ is equal to that of the tetrahedron $ABC'D$, since the two tetrahedra have the same base $ABD$ and the vertices $C$, $C'$ lie on a line parallel to the base plane. In the same way one sees that the tetrahedra $ABC'D$ and $ABC'D'$ have the same volume, having the same base $ABC'$ and vertices $D$, $D'$ on a line parallel to the base plane. It follows that the volume of $ABCD$ is equal to that of $ABC'D'$; since $AB$ is perpendicular to $\pi$, this volume is given by $\frac{1}{3} a S$.