It is not a direct answer to your question, but reading this answer (mentioned in the comments) is an excellent way to understand what is necessary for this circuit to, sort of, work as a memory.
With that understood, you can take the next step to understand that the propagation delay is important to allow the storage. It determines how long the input signal must be strongly pulled to high or low for the output to be able to weakly sustain the input. The input can only turn to "floating" after this delay.
Note that an ideal circuit, with zero propagation delay, would also work, given the conditions presented in the linked answer. A direct answer would then be: propagation delay is not required for the circuit to work as a memory but if it exists, it must be taken into account for the effective storage.