You show E1 to E4 connected to ground instead of the phototransistor emitters.
The optimal LED current will be in the 10 to 20 mA range - as your graph shows. Let's say 20 mA and a 100 Ohm resistor. Worst case CTR is 0.5, so you may expect 10 mA from the phototransistor, at 24V that would correspond to 2k4 and the resistor would dissipate a quarter watt. I think 4k7 would be a more conservative choice (if you needed them - see below).
Do we know yet what the input characteristic of your 24 V device is? Since the datasheet refers only to switches pulling up to 24 V, you do not need pull-down resistors in any case.
Using 43 ohm current limiting resistors will destroy your Pro Micro. You cannot draw 50 ma from an output pin, 20ma is the reasonable, practical limit. 150 ohms, perhaps 120 ohms would be the absolute minimum value to use with a 1.15 vf led.
Without knowing what current is needed to switch your device, it is difficult to know whether it will switch reliably, but it is unlikely that it will require more than 10 mA, so it should be fine.Sadly, "WattsThat" has failed the basic requirements for answering a question, which is to comprehend the question asked by reading it in the first place.
I do not understand why my LED resistor has something to do with the current on the phototransistor side? Because the phototransistor is more activated if the led is brighter?