How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then
$$
|\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y|
$$
so
$$
\rho(\alpha x + \beta y) =
\rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le
\rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y)
$$
So it seems we also need $\rho(|x|) =\rho (x)$.