Wire the LED with another 1N400x and 2.2k/.5w series resistor across the diode in the circuit. The LED will not light, until the original 1N4004 diode fails (open), after which, the current needed for the LED will pass thru it. The only adjustment to this circuit will be to lower the value of the series resistor to increase the brightness of the LED should the circuit load not be large.
It will not detect a shorted diode, only those that open.
Hope this helps.
Jim