Does NaOH react with O3 to produce NaO3?
My own Google search does show that sodium ozonide can form [1].  But it requires low temperature:

Sodium ozonide was produced by the ozonization of sodium hydroxide at -80 to -100°[C] followed by extraction with liquid ammonia and removal of the latter under vacuum at-50° [2]. Composition of the product obtained: $\ce{NaO3}=80.1, \ce{NaOH•H2O}=11.2, \ce{H2O}=8.6$ [%].

Upon warming it decomposes to give the superoxide, and then in the presence of water back to sodium hydroxide:

The process of thermal decomposition of sodium ozonide to $\ce{NaO2}$ occurs at −20 and −10°.


At temperatures above 0°, the process of thermal decomposition of sodium ozonide is accompanied by the process of interaction of the $\ce{NaO2}$ formed with $\ce{H2O}$, and decomposition proceeds to the formation of $\ce{NaOH}$.

References
1.
Tokareva, S.A., Pilipenko, G.P. Thermal decomposition of sodium ozonide. Russ Chem Bull 13, 686–688 (1964). https://doi.org/10.1007/BF00845322.
2.
S. A. Tokareva and M. S. Dobrolyubova, Izv. AN SSSR, Ser. Khim. 1964, No. 4.