This is "too thorough", to put it nicely. One wonders why you just didn't start with $0' + 0 = 0' + 0.$ and conclude $0 = 0'$, using commutativity if you felt necessary.
As another general piece of advice, at the end of each proof by contradiction, look back and see if you can't write it as a straightforward "direct proof" that doesn't use contradiction. This particular proof is a good candidate for that. Unnecessarily using contradiction proofs can lead to proofs that are more complex than necessary.
That said, proofs by contradiction are sometimes the way to go, or else lead you to a method for a direct proof, so they have their uses.
Let us consider the basis of $U \otimes V$. As we are dealing with $\mathbb{R}^3 \otimes \mathbb{R}^3$, the basis $B$ is the set of all $e_i \otimes e_j$ for $i,j \in \{1, 2, 3\}$. Note that for the cross product, the multiplication table is as follows
$$
\begin{array}{c|c|c|c|}
\times & e_1 & e_2 & e_3 \\
\hline
e_1 & 0 & e_3& -e_2\\
\hline
e_2 & -e_3 & 0 & e_1 \\
\hline
e_3 & e_2 & -e_1 & 0 \\
\hline
\end{array}
$$
Then define a map $W: B \to \mathbb{R}^3$ by the following multiplication table
$$
\begin{array}{c|c|c|c|}
W & e_1 & e_2 & e_3 \\
\hline
e_1 & 0 & e_3& -e_2\\
\hline
e_2 & -e_3 & 0 & e_1 \\
\hline
e_3 & e_2 & -e_1 & 0 \\
\hline
\end{array}
$$
As an example of how the table is to be read, this table says that $W(e_1 \otimes e_2) = e_3$.
Now that we have defined $W$ on the basis, we can extend to the rest of $\mathbb{R}^3 \otimes \mathbb{R}^3$. This is because the tensor product and the cross product have the same structure. The cross product is a bilinear map, that is:
$$
\begin{align}
(v_1 + v_2) \times w = v_1 \times w + v_2 \times w \\
v \times (w_1 + w_2) = v \times w_1 + v \times w_2 \\
\end{align}
$$
and the tensor space follows a set of "bilinear" axioms:
$$
\begin{align}
(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w \\
v \otimes (w_1 + w_2) = v \otimes w_1 + v \otimes w_2 \\
\end{align}
$$
These are very similar. Now, extending $W$ linearly, we truly see that
$$W(v \otimes w) = v \times w$$
As an example, which illuminates the proof of this fact, consider the following computation:
$$
\begin{align}
W((1\mathbf{e}_1 + 2\mathbf{e}_3) \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\
W(1\mathbf{e}_1 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3) + 2\mathbf{e}_3 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\
W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2 + 1\mathbf{e}_1 \otimes 4\mathbf{e}_3 + 2\mathbf{e}_3 \otimes 3\mathbf{e}_2 + 2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\
W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2) + W(1\mathbf{e}_1 \otimes 4\mathbf{e}_3) + W(2\mathbf{e}_3 \otimes 3\mathbf{e}_2) + W(2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\
W(3(\mathbf{e}_1 \otimes \mathbf{e}_2)) + W(4(\mathbf{e}_1 \otimes \mathbf{e}_3)) + W(6(\mathbf{e}_3 \otimes \mathbf{e}_2)) + W(8(\mathbf{e}_3 \otimes \mathbf{e}_3)) &= \\
3W(\mathbf{e}_1 \otimes \mathbf{e}_2) + 4W(\mathbf{e}_1 \otimes \mathbf{e}_3) + 6W(\mathbf{e}_3 \otimes \mathbf{e}_2) + 8W(\mathbf{e}_3 \otimes \mathbf{e}_3) &= \\
3\mathbf{e}_3 - 4\mathbf{e}_2 - 6\mathbf{e}_1 + 8\cdot\mathbf{0} &= \\
-6\mathbf{e}_1 - 4\mathbf{e}_2 + 3\mathbf{e}_3
\end{align}
$$
which is indeed the cross product of those two vectors. Steps 4 and 6 rely on the fact $W$ had been extended linearly, while the rest of the steps used the bilinear properties of the tensor product. (Compare this to explicitly finding the cross product by expanding and multiplying!)
Conclusion: The tensor product allows us to use its axioms of bilinearity to "mimic" the properties of a bilinear object or map.
Best Answer
We have that
$$ (-v,w) + (v,w) = (0,w) = (0,0), $$
since
$$ (0,w) = (00,w) = (0,0w) = (0,0) $$
and likewise $(v,0) = (0,0)$. This is not a contradiction because on $V \otimes W$, all these pairs are 'declared' to be equal. Likewise
$$ (0,0) + (v,w) = (v,0) + (v,w) = (v,0+w) = (v,w). $$
A more formal treatment of this could be done via equivalence relations, which I highly encourage you to read on. With this machinery, one takes the cartesian product $V \times W$ and then identifies some pairs as equivalent, which is what 'declaring' equality formally means in the article. Then, $v \otimes w$ is nothing more than the equivalence class of $(v,w)$.