It confused me (and still does) that the board yields desired output when powered by the USB, and fails when powered by an external power supply.
The on-board regulator is a linear regulator. When drawing large amounts of current through it, it gets hot and will start to shut-down. If your LCD was really drawing its 250mA, plus 50mA for the Arduino, that's 300mA. 10V - 5V = 5V drop across the regulator. P = I * E. So Power Dissipated is 300mA * 5V = 1.5Watts. That's a lot of power for the regulator to dissipate without a heat sink.
You should measure the 5V pin (and Vin) when the the external supply is connected. If the 5V pin is well below 5V, you know that the regulator is shutting down.
The good news
is that this is a "good thing(tm)". The regulator is designed to keep itself from overheating, keeping it from getting damaged.
Reading your post makes me realize it doesn't matter whether it works with USB, I shouldn't be driving the backlight with an IO pin.
No, you should not. You should be driving a transistor with the I/O pin and have it turn on the backlight.
If I pull the VIN to a break out board, can I then use that to drive the backlight or is the VIN pin regulated/limited as well?
The only real limitation is how much current the traces on the PCB can handle. If you're drawing more than 1A, then I would rethink just drawing through Vin.
Do I need to setup a voltage regulator (7805) to be driven by the VIN? I think I experimented with this and had a similar failure... powering the backlight from the VIN still caused the arduino to reset.
This depends entirely on how it was wired.