Role of catalyst in elementary reactions towards rate law
The catalyst uses a different reaction path with the zero net catalyst consumption, compared to not catalyzed reaction.
E.g. instead of $\ce{A + B &lt;=&gt; C}$, there is $\ce{A + M &lt;=&gt; AM}$, $\ce{AM + B &lt;=&gt; M + C}$.
If there is a homogenous catalyst, then its concentration in rate equations is processed by the same way as the reactants or products. But the different reaction path, compared to non catalyzed reaction, remains.
So it would not be $$\ce{O3 + M &lt;=&gt; O2 + O + M}$$
but rather e.g. :
$$\ce{O3 + M &lt;=&gt; O2 + MO}$$
and
$$\ce{MO &lt;=&gt; O + M}$$

OR:
$$\ce{O3 + M &lt;=&gt; O + MO2}$$
and
$$\ce{MO2 &lt;=&gt; O2 + M}$$
So you need to solve the kinetics of the system of linked reactions, considering catalyst form concentrations in steady states.

If it is a heterogenous contact catalysis, then the amount and activity of catalyst is implicitly involved in the catalyst-dependent kinetic constant:
$$\frac{\mathrm d[\ce{O2}]}{\mathrm{dt}} = k_{1,\mathrm{M}} [\ce{O3}] - k_{-1,\mathrm{M}}[\ce{[O2][O]}$$
where the ratio $\frac{k_{1,\mathrm{M}}}{k_{-1,\mathrm{M}}}$ remains constant, independent of the catalyst activity.