For all $x,u\in E$ and $y, v\in F$, we have \begin{eqnarray*} \langle A\otimes B(x\otimes y), A\otimes B(u\otimes v)\rangle &=& \langle Ax\otimes By, Au\otimes Bv\rangle\\ &=& \langle Ax, Au\rangle \langle By, Bv\rangle\\ &=& \langle x,u\rangle \langle y,v\rangle\\ &=& \langle x\otimes y, u\otimes v\rangle. \end{eqnarray*} Since vectors of the form $x\otimes y$ or $u\otimes v$ generate a dense subspace of $E\otimes F$, the operator $A\otimes B$ preserves the inner product on the whole space $E\otimes F$. If $E\otimes F$ is infinite-dimensional, we additionally have to show that $A\otimes B$ is surjective (cf. Wikipedia). Let $x\in E$ and $y\in F$, then there exist $u\in E$, $v\in F$ such that $Au=x$ and $Bv=y$ (because $A$ and $B$ are unitary and thus surjective), so $A\otimes B(u\otimes v) = x\otimes y$. Since vectors of the form $x\otimes y$ generate a dense subspace of $E\otimes F$ and $A\otimes B$ is an isometry, we conclude that $A\otimes B$ is surjective.