Why does ozone have higher entropy than oxygen?
You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen,
$\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions. 

Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.

The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have
$$
\Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1}
\quad
\Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1}
$$
and for the standard entropies
$$
S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
\quad
S^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need
$$
\Delta_{\text{f}} S^\circ (\text{O}_3) 
= S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2) 
= -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.

So the reaction has a negative $\Delta S$ and a positive $\Delta G$.



[EDIT following OP comment]

This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates,
but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.

This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in which
the first step is the dissociation of O$_2$ into two oxygen atoms,
by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy. 

Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's Chemistry of Atmospheres give a lot more detail.



[2nd EDIT following further OP comments]

For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here)
$$
\Delta_r G^\circ = -RT \ln K
$$
where $K$ is expressed (for ideal gases) in terms of the partial pressures
relative to the standard pressure $p^\circ$ (1 bar)
$$
K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3}
$$
For large positive $\Delta_r G^\circ$, $K$ is extremely small,
so the equilibrium partial pressure of ozone is extremely small,
if we are given that the partial pressure of oxygen has a reasonable value.
Indeed, if we take (for the reaction as written, producing two moles of ozone)
$$
\Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1}
$$
then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$;
if the partial pressure of oxygen is $\approx 0.21$ bar
(as in the atmosphere around us),
then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.

It is wrong to conclude that the reaction producing ozone "cannot happen"
because the free energy change is positive.
$\Delta_r G^\circ$ is simply the free energy change associated with
the complete conversion of reactants into products.
In reality, an equilibrium is always established somewhere in between.
In this case, it is very far over to one side, but not 100%.
If $\Delta_r G^\circ$ were not so large (but still positive) the
equilibrium would not be so far in favour of reactants.
Thermodynamically, you can think of the position of equilibrium
as being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products:
this will always result in an equilibrium position that is neither 100% reactants
nor 100% products.

It is also possible to give a dynamical interpretation of chemical equilibria
in terms of forward and reverse reaction rates,
but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps
(nobody is pretending that three molecules of oxygen miraculously collide
together at once). To see why this more complicated mechanism
does not affect the argument above about the equilibrium constant, see 
How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium
$$
\text{A} + \text{B} \leftrightharpoons \text{C} + \text{D}
$$
and assume that the forward and reverse reactions are simple one-step processes
obeying the rate equations
$$
\text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}],
\qquad\text{and}\qquad
\text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}].
$$
At equilibrium both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing.
We can write
$$
\text{Rate}(\rightarrow) = \text{Rate}(\leftarrow)
\qquad\Rightarrow\qquad
K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} 
= \frac{k_\rightarrow}{k_\leftarrow}
$$
So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$,
and $k_\leftarrow \gg k_\rightarrow$.
The rate constant for the reverse reaction is much larger than that for the
forward reaction.
However, both reactions are still happening, and the rates of these reactions
are equal.

As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case.
Also, most of the above material is covered in standard physical chemistry texts.