Electronic – SMD resistor maximum voltage: is 200V really a limiting factor

The PO596 is quite small and is suitable for 1400 volts so maybe use 4 in series to give you the 5kV requirement. Clearly, the resistance for each has to drop to 55kohm or thereabouts. All four resistors should be nominally identical in value.

Without the relevant section of EN60664-1 spelled-out anyone reading this question is probably going to be "in the dark" but maybe you have an internet source?

The name of the spec is: -

Insulation coordination for equipment within low-voltage systems. Principles, requirements and tests

Based on this I would say that your interpretation may be flawed - the 220k resistor in series with the opto is, presumably connected AC wire to AC wire (live to neutral) and this has no relevance as "insulation". It isn't an insulator.

the light bulbs are 1.5V @ 0.3A

How do you expect to get 300 mA thru 1 kΩ resistor with only 4.5 V applied? Even with the full 4.5 V across a 1 kΩ resistor, you only get 4.5 mA flowing thru it. With two such resistors in parallel, you get 4.5 mA thru each, so 9 mA total. That's not going to light up a bulb that requires 300 mA for normal brightness. And, this isn't even accounting for the 1.5 V drop across the bulb when it is lit.

One solution is to use lower resistances. Using Ohm's law, we can calculate what they should be. You have 4.5 V available, and the bulb will drop 1.5 V of that when fully lit. That leaves 3.0 V across the resistor. With 3.0 V across it, the resistor should pass 300 mA. (3.0 V)/(300 mA) = 10 Ω. That's the total series resistance for fully lighting the bulb. Since you want that to happen with two resistors in parallel, they should each be 20 Ω.

Also consider the power the resistors will dissipate. Let's be safe and say they need to survive the full 4.5 V applied to them, because stuff inevitably happens. (4.5 V)²/(20 Ω) = 1.0 W. Therefore get "2 W" resistors. Those are also larger and easier to see, so better for demonstration purposes anyway.