Working over vector spaces, let $f:U \longrightarrow U'$, $g: V \longrightarrow V'$. Define $f \otimes g : U \otimes V \longrightarrow U' \otimes V'$ by $(f \otimes g)(u\otimes v) = f(u)\otimes g(v)$.
Then is $f\otimes g$ an isomorphism iff $f$ and $g$ are isomorphisms?
(Really, I'm only asking if the forwards direction is true since it's trivial to show that if $f^{-1}$ and $g^{-1}$ exist, then $f^{-1}\otimes g^{-1} = (f\otimes g)^{-1}$.)
The analogous question with direct sums is easy to prove, but that because it's easy to restrict a map from direct sum to one of its components; I'm having trouble seeing (assuming it's possible) how to define a map $f^{-1}:U' \longrightarrow U$ from $(f \otimes g)^{-1}:U'\otimes V' \longrightarrow U\otimes V$.
Suppose that $U$ and $V$ are non-zero, and that $f\otimes g$ is an isomorphism (and therefore $U'$ and $V'$ are non-zero).
Then both $f$ and $g$ must be injective, because if $f(u)=0$ for non-zero $u\in U$ then for any non-zero $v\in V$ we would have $(f\otimes g)(u\otimes v)=0$.
Now we wish to prove that $f$ and $g$ are surjective. Let $(u'_i)_{i\in I}$ be a basis of $\mathrm{Im}(f)$ and extend to a basis $(u'_i)_{i\in J}$ of $U'$. Let $(v'_k)_{k\in K}$ be a basis of $V'$. Then $U'\otimes V'$ has $(u'_i\otimes v'_k)_{i\in J,k\in K}$ as a basis, but the image of $f\otimes g$ is spanned by $(u'_i\otimes v'_k)_{i\in I,k\in K}$, which is a contradiction unless $I=J$.