Unable to understand the details of step-down voltage regulator
To light a LED when the regulator output is "good", you don't do anything to the EN pin.

When the output is good, the PG pin is high impedance, and when not good, it is pulled low.  This needs to be inverted if you want the LED to light when power is good.

The simplest solution, although hard on the PG output, is to connect a resistor between the PG output and 5 V, and the LED between the PG output and ground.  For a green LED that drops 2.1 V and 10 mA LED current, the resistor would need to be 300 &Omega;.  When the PG output goes low, it sinks the current instead of the LED, so the LED shuts off.  In this example, the PG output would need to be able to sink 17 mA.

Here is a more sophisticated solution:



It still runs the LED at 10 mA when power is good, but puts much less of a load on the PG pin.  It's possible that the PG pin can't sink 17 mA, as required by just the resistor and LED circuit described earlier.  This circuit puts less than 1 mA load on the PG pin, which it almost certainly can do.  It also draws less current from the 5 V supply when the LED is off.  The first circuit actually draws more when the LED is off.