$v \otimes v' = v' \otimes v$ implies $v = av'$

Your approach is good, but your last step is wrong: you can't conclude from the equality $$\sum_{i \in I}r_i e_i \otimes \sum_{j\in I} s_j e_j = \sum_{j\in I} s_j e_j \otimes \sum_{i \in I}r_i e_i$$ that the $(i,j)$ term on one side is equal to the $(i,j)$ term on the other side. Instead, if you write the right side as $\sum_{i\in I} s_i e_i \otimes \sum_{j \in I}r_j e_j$, you can conclude the $(i,j)$ terms on each side are equal, since the elements $e_i\otimes e_j$ form a basis for $V\otimes V$, and so the coefficients of $e_i\otimes e_j$ on each side must be the same. This gives $r_is_j=s_ir_j$ for each $(i,j)$.

From here you could argue that these equations (and the assumption that $v,v'\neq 0$) imply that there is some $a$ such that $r_i=as_i$ for all $i$. However, there's actually a trick you can use to make it a lot easier. Everything we've done is valid for any basis, so let's choose a nice basis. In particular, assuming that $v$ is not a scalar multiple of $v'$ (so they're linearly independent), let's choose a basis with $e_1=v$ and $e_2=v'$. Then in this basis we have $v\otimes v'=e_1\otimes e_2$ and $v'\otimes v=e_2\otimes e_1$. Since these are distinct elements of our basis for $V\otimes V$, they cannot be equal.

If u and v are not linearly dependent, then you can construct a basis which contains them both, and from that a basis of the tensor product. That basis then contains those two elementary tensors that appear in your question, and they are therefore different.