Relativistic transformation of $c^2/v$

Well, for what this is worth:

Fix $w$ and put $f(v)=(v+w)/(1+vw)$. Think of $f$ as a map from ${\mathbb R}^+$ to itself.

(Here ${\mathbb R}^+$ is the positive real numbers. $f$ is also defined for negative reals, but never mind that now.)

What you are saying is equivalent to:

$f$ commutes with the taking of inverses.

We can identify ${\mathbb R}^+$ with ${\mathbb R}$ by applying the log function in one direction and the exp function in the other. These maps are homomorphisms and so take inverses to negatives. After this identification, we have a map $\hat{f}:{\mathbb R}\rightarrow {\mathbb R}$ given explicitly by $x\mapsto \log(f(\exp(x)))$. Under this identification, and accounting for what's been said about homomorphsims, what you are saying is that

$\hat{f}$ commutes with multiplication by $-1$

or in other words

$\hat{f}$ is an odd function.

This is easily confirmed by a direct calculation and/or by an appeal to the calculation in your post.

I'm not sure this adds anything, other than to point out that your observation comes down to the fact that a particular function is odd. Odd functions are not all that rare, and it's not always considered odd that they crop up from time to time, so maybe that's all there is to it.

Rephrasing a bit, the relativistic addition law says that if $v_1$ and $v_2$ sum to $v$, which we'll write as $v_1 \oplus v_2 = v$, then their "boost angles" (or rapidities) add directly. That is, $$\tanh \phi_i = v_i, \quad \tanh \phi = v, \quad \phi_1 + \phi_2 = \phi.$$ Your observation is that $1/v_1 \oplus v_2 = 1/v$ as well, which is equivalent to showing that $$\text{arctanh}(a) - \text{arctanh}(1/a) = \text{constant}$$ for all $a \in (-1, 1)$. And this turns out to be true, so maybe your result just boils down to a coincidental property of arctanh.

I've tried to come up with a physical explanation. Having $|v| > 1$ corresponds to having a complex 'boost angle' $\phi$. I don't know what that means, and it's extra confusing because boost angles are already 'imaginary' generalisations of ordinary rotation angles (i.e. they are what you get if you rotate by an imaginary angle). So maybe $|v| > 1$ wraps it back around to being a normal rotation, but I can't see it.