Let $K$ be a field. Suppose $A$ and $B$ are $K$-algebra and there is a derived equivalence $F:D^b(A)\cong D^b(B)$ between their bounded derived categories. If we assume that $A$ has a tensor decomposition $A\cong A_1\otimes_K A_2$ for two $K$-algebras $A_1$ and $A_2$.
Does $B$ also have a tensor decomposition $B\cong B_1\otimes_K B_2$ such that $D^b(B_1)\cong D^b(A_1)$ and $D^b(B_2)\cong D^b(A_2)$.
Not in general.
For an easy example, let $C$ and $D$ be derived equivalent algebras. Then $A=C\times C$ and $B=C\times D$ are derived equivalent, and $A=C\otimes_K(K\times K)$ has a tensor decomposition, but $B$ will usually not.
There are also connected examples. For example, $KA_2\otimes_K KA_2$ is derived equivalent to $KD_4$ (where $KA_2$ and $KD_4$ are path algebras of quivers of the indicated Dynkin type).
There are more examples in this question. But also, examples are very easy to construct. If you take an algebra of the form $A=A_1\otimes_KA_2$, then it will usually have many tilting complexes, and unless you choose one of the form $T_1\otimes_KT_2$, there is no particular reason to expect its endomorphism algebra (which is derived equivalent to $A$) to have a tensor decomposition.
