I'll only show that it sans $V_1\otimes V_2$, once Adam's proof is wrong in this part.
Consider $U=\operatorname{span}\{e_i\otimes f_j:i\in I,j\in J\}$, with $\{e_i:i\in I\}$ basis for $V_1$ and $\{f_j:j\in J\}$ basis for $V_2$.
We have trivially that $U$ is a subspace of $V_1\otimes V_2$.
Let $h:V\times W \to V\otimes W$ be the bilinear map associated to the tensor product via universal property. That is, $h(v,w)=v\otimes w$.
It is clear that $\operatorname{im}(h)=\operatorname{span}\{h(e_i,fj)=e_i\otimes f_j: i\in I,j\in J\}=U$. But, in this case, $(U,h)$ also satisfies the universal property of tensor product. Thus, $U\cong V_1\otimes V_2$. Then, $U=V_1\otimes V_2$.
Thus, $V_1\otimes V_2$ is generated by $\{e_i\otimes f_j:i\in I,j\in J\}$.
Now, suppose that $\sum_{ij}a_{ij} e_i\otimes f_j=0.$ Then, $0=\sum_{ij}a_{ij} e_i\otimes f_j=\sum_{ij}a_{ij} h(e_i,f_j)$