_I got it meanwhile; answers are still heartly welcome!_
For densely-defined operators it holds: $$A\text{ closable}\iff A^*\text{ densely defined}$$
But one has the inclusion: $$A^*\otimes B^*\subseteq(A\otimes B)^*$$ So the result follows from denseness of the adjoint.
_(Note that this proof works only for densely defined operator.)_