Can a series RLC circuit have an unstable natural response?
This is an algebraic error of the author. Look at the proper step that the author should have done in the "solving we have" part,

$$LC(s^2V_C(s)-sV_0)+RC(sV_C(s)-V_0)+V_C(s) = 0,$$

$$LC(s^2+sRC+1)V_C(s)-sLCV_0-RCV_0= 0,$$

$$LC(s^2+sRC+1)V_C(s) = sV_0+RCV_0 = (sLC+RC)V_0,$$

$$V_C(s) = \frac{sLC+RC}{LC(s^2+sRC+1)}V_0 = \frac{sLC+RC}{s^2+sRC+1} \frac{V_0}{LC}.$$

Using that \$R = 2\$, \$C = \tfrac{1}{2}\$, \$L = 4\$ and \$ V_0 = 2\$

$$V_C(s) = \frac{2s+1}{s^2+s+1} \frac{V_0}{2} = \frac{2s+1}{s^2+s+1}.$$

As you can see, the initial conditions cannot change the dynamics (poles) of a linear system. Unlike what your book states, the poles will be at

$$s_0 = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm j\sqrt{3}}{2},$$

and the system is stable and the oscillations will eventually decay.

ps: 

Whoever wrote this got the result that a passive system was "unstable" and glanced at the fact that
\$ \lambda(s) = LCs^2+(RC-LCv(0))s+1 \$, meaning, the linear system was not linear (for different starting conditions the "poles" would change, and for nonlinear system the whole idea of poles does not exist/work).