Excellent question! To understand the reason for this, neither expectation of $S_x$ or $S_y$ going to zero is sufficient. One could setup equations from relevant commutation relations and get the probabilities but that’s equivalent to doing the matrix algebra. Let us see if symmetry helps simplifying things.
The fact that $\langle{S_x}\rangle=0$ enforces the weightage between the plus and minus states along x to be equal. This means they are of the general (real) form as follows:
$$|{+,z}\rangle=\alpha~ |{+,x}\rangle ~+~\sqrt{1-2\alpha^2} ~|{0,x}\rangle ~+~\alpha ~ |{-,x}\rangle \\
|{-,z}\rangle=\beta~ |{+,x}\rangle ~+~\sqrt{1-2\beta^2} ~|{0,x}\rangle ~+~\beta~ |{-,x}\rangle $$
But the symmetry of the problem dictates that if you flip the system by $180^\text o$, you should get the same probabilities. In other words, keeping things real,
$$\alpha=\pm \beta$$
Finally, using the fact that the plus and the minus states along z are orthogonal, we get,
$$-2\alpha^2 + 1-2\alpha^2=0\\
\Rightarrow \alpha^2=\frac{1}{4} $$
Where we have chosen $\alpha=-\beta$ as plus would imply both the states are equal.
As you can see this is, like many things in physics, ultimately an outcome of symmetry.
The evolution for a particle with a well defined spin along the $z$-axis is :
$$\begin{array}c
I && III && V \\
|\phi_I\rangle\otimes|+\rangle_z & \longrightarrow & |\psi_\uparrow\rangle\otimes |+\rangle_z & \longrightarrow & |\psi_V\rangle\otimes|+\rangle_z \\
|\phi_I\rangle\otimes|-\rangle_z & \longrightarrow & |\psi_\downarrow\rangle\otimes |-\rangle_z & \longrightarrow & |\psi_V\rangle\otimes|-\rangle_z \\
\end{array}$$
where $|\phi_I\rangle, |\phi_V\rangle$ are wave-packets localized in region I and V respectively, and $|\psi_\uparrow\rangle$, $|\psi_\downarrow\rangle$ are localized in region III on the up and down path respectively.
In the usual Stern-Gerlach experiment, the position of the particle in region III is measured, which allows to tell $|\psi_\uparrow\rangle$ from $|\psi_\downarrow\rangle$ and in turns gives us the spin of the particle.
Now, consider any particle at the start of the experiment. Its state is :
$$|\Psi_I\rangle = |\phi_I\rangle \otimes |u\rangle$$
where $|u\rangle =\alpha |+\rangle_z + \beta|-\rangle_z$ with $|\alpha|^2 +|\beta|^2= 1$. By linearity, it evolves as :
$$\begin{array}c
I && III && V \\
|\phi_I\rangle \otimes |u\rangle & \longrightarrow & \alpha |\psi_\uparrow\rangle\otimes |+\rangle_z + \beta |\psi_\downarrow\rangle\otimes |-\rangle_z & \longrightarrow &|\psi_V\rangle\otimes |u\rangle
\end{array}$$
ie, the spin of the particle is unchanged. Looking only at $III\rightarrow V$, the superposition interferes just in the right way to give the initial spin state.
(In the book, they choose $\alpha = \beta = 1/\sqrt 2$, ie $|u\rangle = |+\rangle_x$, but this works for any initial spin state)
Best Answer
Not surprisingly, such questions date back to the original Stern-Gerlach experiments. An early experiment, perhaps not exactly what you seek, is reported by TE Phipps and O Stern (yes, the same Stern) in Zeitschrift fur Physik 73(3-4) 185-191 (1932). Another, more recent article, is K. Brodsky et al., Europhysics Letters 44(2) 137-143 (1998) on 'Single and double interaction zone with comoving fields in Stern-Gerlach atom interferometry' - this is probably way more sophisticated that you seek, but shows that double Stern-Gerlach experiments are not uncommon, but perhaps not called that.
A nice article that covers most of the related theory in an accessible way is AR Mackintosh, 'The Stern-Gerlach experiment, electron spin and intermediate quantum mechanics', Eur. J. Phys 4 97-106 (1983). This is theory only.