I gave a little more thought to your question last night and, if you want to use your 5VDC adaptor, there is an easy way still using a DC jack. Excuse the scribbled drawing but here's the way to do it. The DC jack has a breaker switch built in between pins 1 & 2 on the drawing. With no plug inserted into the jack, pins 1 & 2 are shorted. When a plug is inserted, it opens the connection between these pins and the barrel of the plug (usually the negative pole) will be in contact with one of these two pins. Be sure and check which pin is in contact with the barrel when you insert the plug into the jack since not all jacks are constructed the same - depends on type and manufacturer. Just make sure you get one with an internal switch.
If you want to use a 9VDC adaptor, just move the positive connection from the DC jack from the 5V output of the 7805 on your circuit to the same point that the positive pole of the 9V battery connected.
In effect what you are doing is connecting one negative lead or the other, depending on which power source you are using, to ground. With the adaptor plugged in, it removes the 9V battery completely from the circuit.
So, hope that helps you with your project.
Mike