Proving the hyponormality of $A\otimes B$

Let $$E$$ be a complex Hilbert space. We recall that an operator $$T\in\mathcal{L}(E)$$ is said to be hyponormal if $$[T^*, T]\geq 0$$ (i.e. $$\langle (T^*T-TT^*)x,x \rangle\geq 0$$ for all $$x\in E$$). Let $$E\overline{\otimes}E$$ denotes the completion, endowed with a reasonable uniform cross-norm, of the algebraic tensor product $$E\otimes E$$.

The pair $$A=(A_1,A_2)\in\mathcal{L}(E)^2$$ is called hyponormal if $$\varphi(A)=\begin{pmatrix}[A_1^*, A_1] & [A_2^*,A_1]\\ [A_1^*, A_2 ]& [A_2^*, A_2] \end{pmatrix}$$ is positive on $$E\oplus E$$ (i.e. $$\langle \varphi(A)x,x \rangle\geq 0$$ for all $$x\in E\oplus E$$.

For $$A=(A_1,A_2)\in\mathcal{L}(E)^2$$ and $$B=(B_1,B_2)\in\mathcal{L}(E)^2$$ we consider $$A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$$. Assume that $$A,B\in\mathcal{L}(E)^2$$ are hyponormal. Why $$A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2$$ is hyponormal?

I think the if part'' is correct and it follows from the following fact: if the operator-matrices $$T = \begin{bmatrix} T_{11} & T_{12}\\ T_{21} & T_{22} \end{bmatrix},\quad S = \begin{bmatrix} S_{11} & S_{12}\\ S_{21} & S_{22} \end{bmatrix}$$ are both positive, then the matrix obtained by entry-wise tensor $$U = \begin{bmatrix} T_{11}\otimes S_{11} & T_{12}\otimes S_{12}\\ T_{21}\otimes S_{21} & T_{22}\otimes S_{22} \end{bmatrix}$$ is also positive. This is similar to Hadamard product of two positive matrices.
On the other hand, there seems to be an issue with the only if part'' (even in the case of single operators): what if you take $A=(0,0)$ and $B$ is arbitrary?