Can a Tensor Product be proved by carefully defining the tensor product between terms? Assume I want to prove something like $\{1,x,x^2\}\otimes\{1,y\}\cong\{1,y,x,xy,x^2,x^2y\}$ where these represent the basis for sets in a polynomial vector space (admittedly, a bit lazily written but you get what I mean). The actual tensor product yields$\{1,x,x^2\}\otimes\{1,y\}=\{1\otimes 1,1\otimes y,x\otimes 1, x\otimes y,x^2 \otimes 1,x^2 \otimes y\}$. If I let $\otimes \mapsto \times$, then it quickly follows that the two basis form isomorphic vector spaces.
I understand the question standing alone sounds naive but my question is more so to ask, what sort of properties does $\times$ have to follow so this sort of substitution constitutes a proof for isomorphism? 
(Note: I have no formal training in this subject, please direct me to another question if it answers this!)
 A: The operation $\times $ has to be bilinear.
Recall the universal property of the tensor product of two spaces $V$ and $W$. The vector space $V \otimes W$ equipped with a bilinear map $\otimes : V \times V \to V \otimes W$ is a tensor product of $V$ and $W$ if for every other vector space $Z$ and a bilinear map $f : V\times W \to Z$ there exists a unique linear map $\overline{f} : V \otimes W\to Z$ such that $\overline{f} \circ \otimes = f$.
Let $\{e_i \otimes f_j\}$ be a basis for $V \otimes W$. If your new operation $\times : V \times W \to V \otimes W$ is bilinear, then you can directly verify that the space spanned by the new basis $\{e_i \times f_j\}$ is also a tensor product of $V$ and $W$. The proof is exactly the same as for $\otimes$ because it uses only bilinearity. Since any object defined by a universal construction is unique up to isomorphism, the two tensor products are isomorphic.
Note: If you are not interested in meanigfully expanding the new operation $\times$ to the entire space $V \times W$, then $\times $ can be anything, really. It's then just a symbol for new basis elements, and the two spaces are still trivially isomorphic since their dimensions are equal. With polynomials, the multiplication operation is defined on the entire space $\operatorname{span}\{1, x, x^2\} \times \operatorname{span}\{1, y\}$, and it is bilinear so it can serve as the bilinear map $\otimes$ from the universal property of the tensor product.
