The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules
If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4).
Over a field, all modules are projective since they're free. So the answer is ‘yes’.
In the present case, here is a sketch of the proof:
Let $K$ be the base field. As $V_1\simeq K^m$ for some $m>0$ and similarly $V_2\simeq K^n$, and the $\operatorname{Hom}$ and $\,\otimes\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove:
$$\widehat\Psi\colon\operatorname{Hom}(K,W_1)\otimes\operatorname{Hom}(K,W_2)\to\operatorname{Hom}(K\otimes K,W_1\otimes W_2)$$
is an isomorphism.
This results basically that for any vector space $V$ we have an isomorphism
\begin{align*}
\operatorname{Hom}_K(K,V)&\simeq V\\
\phi&\mapsto\phi(1)
\end{align*}
In detail, just consider the following commutative diagram:
Your reasoning is not necessarily incorrect, but the issue you see can be mitigated. To do this, we have to slightly ponder what "kind of object" exterior algebras are and what the symbol $\otimes$ means. Namely, if $V$ is a $k$-vector space, then $\bigwedge V$ is not just a $k$-algebra, but a graded $k$-algebra. In fact, it is a graded commutative $k$-algebra (the Wikipedia article calls this "anticommutative" instead - I personally disagree with that choice of terminology), i.e. if $\omega,\eta\in\bigwedge V$ are elements of pure degree, then $\omega\cdot\eta=(-1)^{|\omega||\eta|}\eta\cdot\omega$ (the multiplication here is the exterior product, I use bars to denote the degree).
Now, in general, if you have two graded $k$-algebras $R=\bigoplus_{i\ge0}R_i$ and $S=\bigoplus_{j\ge0}S_j$, then their tensor product $R\otimes_kS=\bigoplus_{k\ge0}\bigoplus_{i+j=k}R_i\otimes_kS_j$ is naturally a graded $k$-vector space with the indicated grading. In fact, $R\otimes_kS$ can be naturally made into a graded $k$-algebra itself, but here is the surprise. If $r\in R_i,s\in S_j,r^{\prime}\in R_{i^{\prime}}$ and $s^{\prime}\in S_{j^{\prime}}$, we ought to define $(r\otimes s)\cdot(r^{\prime}\otimes s^{\prime})=(-1)^{ji^{\prime}}(r\cdot r^{\prime})\otimes(s\cdot s^{\prime})$. This is in accordance with a general principle known as the "Koszul sign rule", which says that when you exchange two graded symbols of degree $j$ and $i^{\prime}$ ($s$ and $r^{\prime}$ here), the sign $(-1)^{ji^{\prime}}$ ought to be inserted. This makes $R\otimes_kS$ into a graded $k$-algebra, which we call the graded tensor product of $R$ and $S$.
Here's a concrete reason why this is the "right" definition: If $R$ and $S$ are graded commutative, then $R\otimes_kS$ is graded commutative too and it would not be if you omitted the sign (check this). Next, if $T$ is any graded commutative $k$-algebra, the two natural $k$-algebra homomorphisms $R\rightarrow R\otimes_kS,\,r\mapsto r\otimes1$ and $S\rightarrow R\otimes_kS,\,s\mapsto1\otimes s$ induce a bijection between pairs of graded $k$-algebra homomorphisms $R\rightarrow T$ and $S\rightarrow T$ and $k$-algebra homomorphisms $R\otimes_kS\rightarrow T$ in the way one would expect (check this and you'll see the sign is crucial here as well). In other words, the graded tensor product is the coproduct in the category of graded commutative $k$-algebras. This generalizes how the (ungraded) tensor product works to the graded setting.
If we now return to the scenario at hand, this deliberation tells us that $\bigwedge V\otimes_k\bigwedge W$ ought to be understood not as a tensor product of algebras, but as a graded tensor product of graded $k$-algebras. The additional sign in the definition of the multiplication in a graded tensor product is precisely the sign that is missing in the equations you write down, so the natural isomorphism $\bigwedge V\otimes_k\bigwedge W\rightarrow\bigwedge(V\oplus W)$ does in fact become an isomorphism of graded commutative $k$-algebras.
Of course, I could've just directly told you to add a sign to the multiplication in $\bigwedge V\otimes_k\bigwedge W$ that magically makes everything work, but I hope this explanation, even if it may not be entirely transparent, highlights that this is not some ad hoc fix, but fundamentally the right perspective on the issue.
Best Answer
Here is an abstract approach. There is no need to assume that $V$ is finite-dimensional or over $\mathbb{C}$: the discussion below applies to $V$ an arbitrary module over an arbitrary commutative ring $R$. The symmetric powers $S^n(V) = V^{\otimes n} / S_n$ organize into a single object called the symmetric algebra
$$S(V) = \bigoplus_{n \ge 0} S^n(V).$$
The symmetric algebra is the free commutative algebra on $V$; formally, it's the left adjoint of the forgetful functor from commutative $R$-algebras to $R$-modules. As a left adjoint, it preserves colimits, so in particular it sends coproducts to coproducts. The coproduct of commutative $R$-algebras is the tensor product, so this gives the "exponential law"
$$S(V \oplus W) \cong S(V) \otimes S(W)$$
and writing this isomorphism down in each degree we get a family of isomorphisms
$$S^n(V \oplus W) \cong \bigoplus_{i+j=n} S^i(V) \otimes S^i(W).$$
The desired isomorphism is the special case $n = 2$. A nice exercise here is to check that when $V, W$ are finite-dimensional vector spaces the dimensions of the two sides agree, which gives a nice identity; more ambitiously you can check that the characters of the two sides agree as representations of $GL(V) \times GL(W)$.