AC sensors (Specially the 2-wire AC sensors) have significant leakage current. this is more than needed to fool your circuit.
the optocouplers are very sensitive, the input LED activates with small current when Vf is very low (about 1.3V).
that is a very low threshold to overcome. you can increase it by adding pair of anti-series zener diodes before optocouplers (at the point A).
the other possibilities are to reduce 1k resistor values for example.
basically your 1k resistors are in parallel (if we ignore optocouples since they are low Vf anyway).
that means their equivalent resistance is about 500 Ohm.
Xc of the 330nF is about 10k.
so those10k and 500 Ohm are forming what is known as voltage divider.
if you reduce value of the 500 Ohm side, your circuit will be less sensitive. in other words it may work for you even though the optos threshold is still very low.

 

the other possibilities are to reduce 1k resistor values

Firstly, thanks for the detailed explanation.
You said that 10k of 300nF + 2 x (parallel 1K) form a voltage divider. Understood, but that divided voltage is not connected to anything. It is simply a junction of 330nF and the 2 resistors. Please correct me if I am wrong.
Next, you want me to reduce the values of both resistors. What target value to achieve?

 

you can increase it by adding pair of anti-series zener diodes before optocouplers

Should 2 x 3.9V zeners do the magic?

 

Changing the value of the vertical 1K resistor to 470Ω did the job.

Thank you for getting me out severe frustration.

 

great... good luck.