Let $k$ be an algebraic closed field with $\mathrm{char}{k}=0$. Assume $A$ is a finite dimensional commutative algebra grading by $\mathbb{Z}_2$, "strongly" means that $A_1A_1=A_0$.
Claim:$A\cong A_0\otimes k\mathbb{Z}_2$ as $\mathbb{Z}_2$ graded algebras.
I think the claim is right and the key is how to find a homogeneous element $g\in A_1$ such that $g^2=1$.
Since $A/J^g(A)$ is gr-semisimple, where $J^g(A)$ denotes the graded radical, $A/J^g(A)$ is a product of some $k\mathbb{Z}_2$. Therefore, we can find $\overline{g}\in (A/J^g(A))_1$, is it liftable?
Assume $(J^g(A))^2=0$, and $g^2=1+r$ for some $r\in J^g(A)$, then $[g(1-r/2)]^2=(1+r)(1-r)=1$. But $g(1-r/2)$ seems not homogeneous?