I removed the LED and I was getting the full 3V, 5V or 8V (whatever Voltage I inputted to the OpAmp) on the output wires with my Multimeter. As soon as I connect the LED, it drops to 1.6V.
An LED acts as a forward biased diode with a forward drop of ~2V. Like a diode, once it begins conducting, it has a very low resistance. If connected to a voltage, some else must be there to limit the current that flows through the LED.
Most LEDs have a maximum current rating of ~30mA, and they are damaged if this current is exceeded by much.
I simulated two identical LEDs, one with a resistor in series with it, the other not. The simulation shows the current through the two LEDs as the input voltage is swept from 0V to 10V (the horizontal X-axis).
Look at the upper plot. No current flows through R1 (or D2) until the input exceeds ~1.8V, at which point the current I(R1) increases linearly to 33mA when the input is 10V. Note that the voltage across D2 V(D) saturates at about 1.8V even though the input voltage goes all the way to 10V. This is the correct way to drive an LED.
Now look at the lower plot. Note that the current I(D1) starts increasing as the input V(in) exceeds 1.8V. By the time the input voltage reaches 3V, the current I(D1) is already 200mA, which is SIX TIMES the max. allowed. This would destroy the LED! Note that even if the LED took it, by the time in the input gets to 10V, the LED current is up to 1.3+A!!!
The only reason you didn't vaporize your LED is because on a good day, the 741 can only put out about 25mA, so it is like it has a built-in current limiting resistor. When you measure the voltage across the LED at the output pin of the 741, it shows the 1.8V just like the upper plot.
Clear as Mud?