Two power supplies for an Arduino Mega - How to handle Vin pin?
Look at the Arduino Mega schematics, there is a circuit that selects either VIN or USB VBUS as the voltage source. It uses LM358 as a comparator, comparing VIN/2 to 3.3V.

So, basically, if VIN is greater than 6.6V, it disconnects USB VBUS from the supply circuit and it is then safe to use VIN.

So you can use the circuit you thought of (the diode isn't even strictly necessary, but is certainly safer), and it won't put 2.8V across the diode, because USB VBUS won't be connected to VIN, even when both the DC jack and the USB are connected. Everything will work as you expect.

Note that it works fine because 7.5V minus the diode drop is greater than 6.6V. Otherwise, you'd need to use a greater supply voltage. Also, if I were you, I'd use a schottky diode, not a silicon one.