As Mikael Patel pointed out in his answer, you have set up a voltage divider. Mikael has thoughtfully provided a schematic of the clone Uno, where the relevant part is here:
If you jumper together pins D0 and D1 (Rx and Tx) on the Uno effectively you have made this circuit:
If I set up this with a function generator, I get this result:
You can see that for a square-wave input of 0 to 5V, we only get an output of 2.16 to 5V.
What is LOW?
Now looking at the datasheet for the DC characteristics we see:
For an input to register as LOW it has to be no higher than
0.3 * Vcc = 0.3 * 5. Thus the voltage cannot be higher than 1.5V. Since we are measuring 2.16V, that will not count as a LOW.
Now admittedly that is for the Atmega328P and not the CH340G, but it will probably have similar characteristics.
I found a datasheet for the CH340G, and that shows even stricter requirements for a LOW signal:
Comparing to the official Uno circuit:
You will note there that the indicator LEDs are on a different circuit, and thus do not form a voltage divider.
The clone Uno circuit above can be re-arranged like this:
Given that the LEDs have a voltage drop of 1.86V (measured in practice, and about right for the red LEDs I am using) then effectively we have the resistors R1 and R2 connected to 3.14V.
Since they are in parallel they are effectively 500 ohms.
Now using the formula for voltage dividers:
1000 / (1000 + 500) * (5 - 1.86) = 2.09V
This is close enough to the observed result. I didn't check each resistor to make sure it was exactly in spec.
I don't think R4 affects the results because it is effectively "floating".