$f \star (g \,* \,h) = (f \star g) * h$?
It's easiest to look at this in the frequency domain. We have
$$\mathcal{F} (x*y) = X \cdot Y$$
$$\mathcal{F} (x \star y) = X' \cdot Y$$
where ' is the complex conjugate
So we get
$$\mathcal{F} (f \star (g * h)) = F' \cdot G \cdot H $$
$$\mathcal{F} ((f \star g) * h) = (F \cdot G)' \cdot H = F' \cdot G' \cdot H  $$
These are only identical if G is real.
In your case you let's simplify the notation to $f = S_0, g = S_0, h = H$ $S_0$ has a unit impulse as autocorrelation, which means that the spectrum of the autocorrelation is just unity, i.e. $1$
$$\mathcal{F} (f \star f) = \mathcal{F} (\delta) = 1 = F' \cdot F$$
We use that property and can show that
$$\mathcal{F} (f \star (f * h)) = F' \cdot F \cdot H = H$$
since $F' \cdot F = 1$