Electronic – Why doesn’t the multi-meter diode tester work on this LED

There are various misconceptions here.

The emitter resistor value has no effect on efficiency. All the current for each LED is coming from the 5V supply. Whatever part of that voltage the LED doesn't use times the current is going to be wasted as heat. Small emitter resistors only more the dissipation to the transistors. The total dissipation is the same.

I thought I mentioned it in your other question, but the emitter resistors should be a larger value. At 1 Ohm just 1 1mV offset will cause 1mA thru the LED, which is probably dim but visible. From your voltage divider ratio and the 1 Ohm emitter resistors, it looks like you're aiming at a bit over 300mA LED current. Didn't I go thru a detailed calculation of the emitter resistor value in your other question?

I don't know what kind of LEDs these are, but most likely you can afford at least a volt accross the emitter resistor, so 3.3 Ohms would be a better choice that will give you more control.

As for the not quite 0 and 5 volt output, that's probably something the Arduino is doing. Keep in mind that arduinos are Simplified for the masses. It wouldn't surprise me if there is a resistor in series with each output as protection.

The opamps are probably working fine enough. Every opamp has some offset error, and if these are a little positive, their outputs will go high just enough to turn the transistors on a little to make that offset voltage appear accross the emitter resistor. This is yet another reason for using a larger emitter resistor.

The 2 rails at the top of a breadboard are all connected along the straight line. In the picture, you are connecting both ends of the resistor to the same rail, so it's as if it's not even there (all the current will bypass it and travel through the LED).

It's the equivalent circuit of having the resistor ends connected (same voltage):

          +---R---+      <----- Resistor is useless, current will bypass it
          |       |
 +--------+-------+-----+
 |                      |
(+)                     |
 V                    (LED)
(-)                     |
 |                      |
 +----------------------+

If you are trying to have an LED to let you know when your power is connected, you will either need to use a pin line outside of the voltage rails (and make sure the current passes through the resistor), or else use an LED with a built-in resistor. As your circuit is now though, you run the risk of shorting out your battery, depending on how much current the LED allows through.