I apologize for that the following answer may result from a total misunderstanding of the problem. .

Let $X=M\times N$, and let $E=\pi_M^*TM$ and $F=\pi_N^*TN$. Note that $TX=E\oplus F$. As suggested in Ben McKay's answer, you could first look for the existence of an almost complex structure $J\in End(TX)$, $J^2=-Id$, for which the eigen-bundles of $\pm i$ are precisely $E^{\mathbb{C}}$ and $F^{\mathbb{C}}$.

This may exist or not. It does for instance under the assumption that there exists an isomorphism of vector bundles $\phi:E\to F$. In that case, define $$ J:=\left(\matrix{0 &\phi^{-1} \\ -\phi & 0}\right) $$ In this situation it seems to me that the almost complex structure is automatically integrable: indeed, sections of $E^{\mathbb{C}}=\pi_M^*TM^{\mathbb{C}}$ are stable under the Lie bracket.

[EDIT: I started to write this answer, and then realized it doesn't really answers the question - I leave it here anyway, just in case]

Let $X=M\times N$, and let $E=\pi_M^*TM$ and $F=\pi_N^*TN$. Note that $TX=E\oplus F$. As suggested in Ben McKay's answer, you could first look for the existence of an almost complex structure $J\in End(TX)$, $J^2=-Id$, for which both projections $\pi_M$ and $\pi_N$ have totally real fibers.

This may exist or not. It does for instance under the assumption that there exists an isomorphism of vector bundles $\phi:E\to F$. In that case, define $$ J:=\left(\matrix{0 &\phi^{-1} \\ -\phi & 0}\right) $$ In this situation the eigenbundle $T^{1,0}X$ corresponding to $+i$ is isomorphic to $E^{\mathbb{C}}$ in the following way: $$ E^{\mathbb{C}}\to T^{1,0}X\quad;\quad v\mapsto v+i\phi(v) $$ You may want to look for conditions on $\phi$ for $J$ to be integrable.