Why Fourier transform and Stockwell-transform retain the absolute phase information of one signal?
It simply means that for a signal:
$$
x(t) = a(t) e^{j\phi(t)}
$$
that the phase information $\phi(t)$ is completely preserved in the transform 
domain.

Time-frequency representations that use a quadratic form do not necessarily have this property:
$$
x(t+\tau)x^*(t-\tau) = a(t+\tau)a(t-\tau)e^{j(\phi(t+\tau) - \phi(t-\tau))}
$$
so if $\phi$ is of the form:
$$
\phi(t) = f(t) + C
$$
where $C$ is a constant, then $C$ is not preserved in $x(t+\tau)x^*(t-\tau)$.

This is important because sometimes you are interested in the absolute phase of a signal, not just its relative phase.