Skazzyyy:
I am powering breadboard supply using a 2.1mm DC power cable that is plugged into a wall through a USB power adapter.
So you use a 5volt cellphone charger on the DC socket.
The DC socket might connect to a reverse protection diode that drops ~0.4 or 0.7volt, and a 5volt regulator that drops another volt.
If so, then there is about 3.5volt left on the 5volt output of your breadboard supply.
Measure it.
Skazzyyy:
This may be a newbie question (since I am one for electronics) : I do not understand why I would strictly need <=6volt if the breadboard supply outputs 3.3/5volt ?
Looking at the specs I believe that I would not be able to power it with <=6.5v since these are the specs : "Input voltage: 6.5-9v (DC) via 5.5mm x 2.1mm plug"
Thermal problems.
With 9volt on the DC socket, and a current draw of 500mA from the 5volt rail of the breadboard,
9 - 5 = 4volt * 0.5Amp = 2watt is dissipated in the 5volt regulator of the breadboard.
Ok for a very short while, but too much long term.
A lower input voltage will minimise power (heat) in the regulator.
You might indeed need a minimum of 6.5volt if there is a reverse protection diode inline.
But why don't you connect that 5volt cellphone supply directly to the breadboard/Uno.
Then you don't need that breadboard supply.
Leo..