Given that $u+w+v = 0$, I was able to prove that $u \times v = v \times w = w \times u$ by using the anti-commutative property.
But I'm struggling a lot with how to approach to prove the converse. Given $u,v,w$ are pairwise non-parallel vectors in $\mathbb{R}^{3}$.
On
This is a slightly different angle from Robert's answer. Because $\mathbf{w}\perp\mathbf{s}\times\mathbf{w}$ for all $\mathbf{s}$ nonzero and non-parallel to $\mathbf{w}$ and the vectors are in $\mathbb{R}^{3}$, we have:
$$ \mathbf{w}\perp \mathbf{u}\times \mathbf{v} \implies \mathbf{w}=a\cdot \mathbf{u} + b\cdot \mathbf{v} $$
The first and second equation:
$$ \begin{aligned} \mathbf{u}\times\mathbf{v} &=\mathbf{v}\times\mathbf{w}\\ &=\mathbf{v}\times\left(a\cdot\mathbf{u}+b\cdot\mathbf{v}\right)\\ &=a\cdot\mathbf{v}\times\mathbf{u}+b\cdot\mathbf{v}\times\mathbf{v}\\ &=a\cdot\mathbf{v}\times\mathbf{u} \end{aligned} $$
from which we obtain $a=-1$. Repeating for the first and third equation:
$$ \mathbf{u}\times\mathbf{v}=b\cdot\mathbf{v}\times\mathbf{u} $$
from which we obtain $b=-1$. Therefore, we have
$$ \mathbf{w}=-\mathbf{u}-\mathbf{v} $$
It isn't true. Try $u = v = w \ne 0$.
EDIT: If you add the extra condition that the vectors are pairwise non-collinear, it is true.
Note that $a \times b = 0$ if and only if $a$ and $b$ are collinear.
The condition $u \times v = v \times w$ is equivalent to $(u + w) \times v = 0$, which means $u - w$ and $v$ are collinear. Note that $v \ne 0$ (because any vector and $0$ are collinear), so this implies $u + w = s v$ for some scalar $s$. Similarly, $v \times w = w \times u$ implies $u + v = t w$ for some scalar $t$. Then $u + v + w = (t+1) w = (s+1) v$. The only way this doesn't say $w$ and $v$ are collinear is if $t+1$ or $s+1$ is $0$, and then $u + v + w = 0$.