Electronic – LED as Zener: is this a good way to lower the BOM cost

This should work: -

Rset and Vset determine the current that flows through the load resistor: -

\$I_{LOAD} = \dfrac{V_{SET}}{R_{SET}}\$

However, the op-amp needs to be able to work down to 1.8 volts or lower and will need power connections (not shown above) to V and ground. It also must have rail-to-rail inputs and also preferably outputs.

If Vset is made to be 0.15 volts and Iset is required to be 15mA then Rset is 10 ohms.

The zener diode you reference has a zener-voltage of 5.1V and a power rating of 1W. You want to place a resistor in series with the 9V source and the zener diode to limit current draw.

The datasheet for the IN4733 states that at a zener current of 49mA the zener voltage is at the rated 5.1V. The only calculation you need to do is to size the resistor such that you have at least 49mA, although lower current would also likely give you around the 5.1V zener rating. Normally you would look at the datasheet for a chart to give you this information, but the one I found left this information out.

$$ R = \frac{9-5.1}{10mA} = 390 \Omega$$

But, again, larger sized resistors are likely to give you the voltage you want with lower power dissipation. Test to verify.