Let $E$ be a complex Hilbert space. Let $E\overline{\otimes}E$ denotes the completion,
endowed with a reasonable uniform cross-norm of the algebraic tensor product $E\otimes E$.



> 
> **Definition:** Let $A,B\in \mathcal{L}(E)$, the tensor sum of $A$ and $B$ is defined by 
>  $$A\oplus B:=(A\otimes I)+(I\otimes B)\in \mathcal{L}(E\overline{\otimes}E).$$
> 
> 
> 


Clearly, if $A$ and $B$ are normal operators then so is $A\oplus B$.



> 
> Assume that $A\oplus B$ is normal, it is true that $A$ is normal and $B$ is normal?
> 
> 
> 


A short calculation shows that $A\oplus B$ is normal if and only if 
$$([A^\*,A]\otimes I)+(I\otimes [B^\*,B])=0.$$
It is possible to show that $[A^\*,A]=[B^\*,B]=0$?



Tensor sum of two operators
Let's start with a general lemma. Let $E,F$ be Banach spaces, $E\overline\otimes F$ be the completion of a reasonable crossnorm. Let $\newcommand{\mc}{\mathcal}T\in\mc L(E), S\in\mc L(F)$ be such that $T\otimes I = I\otimes S$. Then $T=S$ is a scalar multiple of the identity.


This follows, as pick $f\in F, f^\*\in F^\*$ with $f^\*(f)=1$. Then
$$ T(e) = f^\*(f) T(e) = f^\*(S(f)) e \qquad (e\in E). $$
Thus $T = \alpha I$ where $\alpha = f^\*(S(f))\in\mathbb C$. Then $\alpha I\otimes I = I\otimes S$ so $S=\alpha I$.


So if your case, we get that $[A^\*,A] = -[B^\*,B]$ is a scalar multiple of the identity. Now use the famous result that the commutant of two bounded operators cannot be a non-zero multiple of the identity. So $A$ and $B$ are normal.


