Since $D \in \mathcal C \otimes \mathcal C,$ it's easy to see that if we define $h:A\times B\rightarrow C \times C $ by $h(a,b)=(f(a),g(b))$, then $h$ is mensurable and since $h^{-1}(D)=E$ and $D$ is mensurable we get that $E$ is mensurable, that is, $E \in \mathcal A \otimes \mathcal B$.