I think this is a very interesting question.
In response to your comment, let me argue that if 1 holds and the measure is additive, then the singleton values are all the same. This is the sense in which the strong form of 2 follows from the weak form of 2.
To see this, following Sean's comment, observe that $\mu (\{a\})+\mu((a,b])=\mu([a,b])=\mu([a,b))+\mu(\{b\})$, and so $\mu(\{a\})=\mu(\{b\})$. So all singletons must have the same measure, and so the strong form of 2 follows from the weak form of 2.
In your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.