Questions about the stability (and stationarity) of a system and state space representations
@w00zie: Hi: I'm putting this here just because I find it easier to write in here. This is not an adequate answer from a DSP point of view (and I would like to see one also) but let me just it explain more clearly since I babbled above and it's still not that clear. 

Take the more standard AR(1):

$y_{t} = a y_{t-1} + \epsilon_{t}$.  ( denote this as first equation )
This can be re-written as

$y_{t}(1 - a L) = \epsilon_t$ where L is the backshift operator which is kind of the analogue to $z^{-1}$ in DSP notation. This expression can be re-written as

$y_{t} = \frac{\epsilon_{t}}{(1 - a L)}$ which can be re-written as

$y_{t} = \sum_{i=0}^{\infty} a^{i} \epsilon_{t-i}$.

But the only way that the infinite sum on the RHS can converge is if $|a| &lt; 1$.
This is why that restriction arises for stability. I hope this helps even though it's still a time-series viewpoint rather than a DSP viewpoint.

Oh, and you asked why it looks so different from the standard AR(1) notationally.

You have: $s_t = 2  s_{t-2} +  3 w_{t-2}$.

If you let $y_t = s_t$ and $\epsilon_t = w_{t-2}$,  then that gives

$y_t = 2 * y_{t-2} +  3 \epsilon_{t}$. ( denote this as last equation ).

The equivalence of "first equation" and l"last equation" can be argued in two different ways: 

1) you can think of last equation as an AR(2) with a zero $y_{t-1}$ coefficient. But the restriction on $a$ for stability doesn't change.

2) you can think of last equation as an AR(1) where the time step is doubled.

Note that the difference in the lags of the error terms, $\epsilon_{t}$ and $w_{t-2}$ is okay because $w_t$ is iid and uncorrelated with all other terms.
You are correct that, if $w_t$ was not iid and uncorrelated with everything else, then you couldn't use 1) or 2) as arguments for equivalence.

Note that I really shouldn't have used $y_{t}$ here since you use $y_{t}$ in a different context but my $y_t$ and your $y_{t}$ are not related.

Finally,  your $y_{t}$ is not stationary (i.e: not stable )  because it's just $s_t$ + noise and $s_t$ is not stationary.