Navigation lamp monitoring circuit
simulate this circuit &ndash; Schematic created using CircuitLab

It reads as though you have created Circuit 1 above. The relay coil will limit the current through the lamp. I think you intended Circuit 2.

Circuit 2 is not much good as it doesn't prove that the light is connected or working.

Please post some details (in your original post) on the LED navigation lamps (current, voltage, wattage) with links to data sheets and we can suggest solutions.

Update after details added to question

Executive summary

Replace your 24 V relay with a 12 V unit. You'll need to add some parallel resistance to pass enough current to light the lamp brightly.

Calculations

We know that LEDs brightness is controlled by current. The fact that the light operates from 9 to 33 V suggests that they've got a current regulator in there. We can figure out what current it's drawing by using the 2.5 W power and the 9 V specification.

Since $$P = V·I$$ we can calculate the current required as $$I  = \frac{P}{V} = \frac{2.5}{9} = 0.28 A$$

Your relay-in-series with lamp approach is the most reliable as if the wires fall off the lamp or it fails open circuit the relay will drop out so let's see if we can make it work.

If we share the 24 V giving half to the lamp and half to the relay we will have the following.







simulate this circuit

We need 0.28 A going through the LED so we need a relay and maybe R3 to pass 0.28 A as well. From Ohm's law we can calculate that the resistance of the coil and R3 in parallel should be
$$R = \frac{V}{I} = \frac{12}{0.28} = 42 Ω$$

Since most suitable 12 V relay coils would have a resistance higher than this we will need R3 in parallel.

$$R3 = \frac{42·Rc}{Rc - 42}$$

Example: 12 V, 400 Ω relay would require
$$R3 = \frac{42·400}{400 - 42} = 46 Ω$$

Finally we need to work out the power rating of R3.

$$P = V·I = 12·0.28 = 3.36 W$$

The nearest stock sizes in this example would be 47 Ω, 4 W wirewound. Note that the resistor will get hot.