You could probably accomplish this kind of charging using a N-type mosfet.
There are still issues with this, however. Like in the circuit you discussed, your voltage output is greatly limited (assuming at typical diode voltage drop of 0.7V, you'll have 3.5V available on adapter power.
In terms of battery power, the best FET I could find might be the
Vishay SUB85N02-03. Accounting only for the threshold voltage you would be limited to a charge of 3.75V, but because of the source-drain voltage drop you'll be limited to a charge of 3V maximum. With the diode taken into account that's 2.3V available for running LEDs at a level where the charge will probably last only a short period of time. Not exactly worth the effort.
Edit: Rerunning numbers now, may have made a mistake:
We know there is a forward voltage drop of 1.2V minimum across the MOSFET. The drain-to-source voltage should be roughly equal to the charger voltage minus the forward voltage minus the battery voltage, and in the orientation this is wired the gate-to-source voltage should be roughly the same. The MOSFET will act as an open switch around the time the gate-to-source voltage equals the threshold voltage, or 0.45V. Knowing this, the maximum battery voltage you can achieve is 2.55V. After the diode you are left with 1.85V left for running your LEDs, and since the lowest forward voltage for a Cree is 2.2V you're kinda screwed in that regard. So in-hilt light up and charge doesn't seem feasible to me.