In a typical 1602 LCD pinout, pin 3 or V0 is the contrast setting for the display. It usually is set to a voltage near ground, often by using a 10K pot with ends connected to Vdd and Vss with wiper to V0.
By putting an LED in series with your resistors (that apparently go to ground) you've either disconnected V0 from ground or have raised its voltage by the LED voltage. This reduces contrast to the point that characters aren't visible even though they are on the display.
The LED backlight remains on and at the same brightness as before because it is driven via pins 15 and 16, independent of all other display settings.
Edit 1: I haven't duplicated your circuit to take measurements, so at the moment don't know for sure; but (in resistors-only case) V0 from pin 3 on the 1602 LCD board apparently sources some small current through your 1500 Ω to ground. When you add an LED it can disconnect V0 from ground or can raise its voltage by the LED voltage, as follows: firstly, if the LED cathode faces V0 and V0 is positive, no current will flow through the diode, thus disconnecting V0. Secondly, if the LED anode faces V0 and V0 is positive, current will flow through the LED. If the LED lights at all, most of its Vf characteristic voltage (typically between 3 and 4 volts for a green LED) will appear across it. Thus, V0 sits at close to Vf plus whatever drops across the resistors.
*The same as for other diodes, LED forward current increases exponentially with voltage (Ref: wikipedia on LED physics) hence voltage change is tiny over large changes in current when an LED is emitting light. In other words, if the LED is lighting up, its voltage will be close (ie within a few dozen millivolts) to its characteristic band gap voltage. See petervis.com's led-resistor-calculator for a "Chart by Colour" that shows characteristic Vf forward voltages.