how does single supply op-amp work while its non-inverting input is connected to ground?
I've copied the relevant figure from the data sheet so that I can speak to it.

The HCNR200 couples the LED output equally to the photodiodes.  So assuming good matching, the current induced on the photodiodes is equal if their voltages are equal.

A photodiode is a current generator, so it will drive its cathode negative when light is shined on it.  So in the circuit to the left, \$V_{IN}\$ tends to drive the op-amp positive input high, which lights the LED.  This generates a current in PD1 which drives its cathode voltage down.  Assuming stability (which you'll find to be a big assumption when you try to make this work), A1 will hold the LED current to whatever it needs to be so that \$V_{IN} - I_{PD1} R1 = 0\$.

This will induce a current in PD2.  Again, the cathode of PD2 will tend to be driven more negative than its anode.  A2 will counteract this by raising \$V_{OUT}\$.  If PD1 and PD2 are nicely matched, then \$V_{OUT}\$ will be proportional to \$V_{IN}\$.