Bypass IC output to give an LED a voltage boost
It may work, but you'll need to experiment and get a bit lucky:

In your second diagram, the effective resistance of the IC + resistor in parallel is going to be less than the resistance of the IC alone, and the smaller your resistance, the more you'll be increasing the voltage drop across the diode. If this is confusing, imagine putting in a resistor that has an extreme resistance:

1) 0 Ohm resistor: This would be like shorting out the path around the IC. All of your 5V drop will be across the LED, and you would probably blow it. Also, all of your current would bypass the IC and the IC would not work.

2) Infinite Ohm resistor: This would be exactly like your first diagram, which is your current situation. The 5V drop is from V(IC) + V(LED).

Somewhere in between these two extremes, you may be able to find a resistance that increases V(LED) without popping it and without letting V(IC) drop too low. You may be able to find it by trying out different resistors. Try starting with a very large resistor and move to progressively smaller resistors until you get the brightness you are looking for. 

Since you can't identify the IC or LED voltage requirements, you're going to have to rely on some luck to find an appropriate resistance. Keep in mind that when an IC's supply voltage drops below the minimum, it may not immediately die. You may end up with behavior that mostly appears normal, yet periodically fails or acts strangely. It may be hard to tell when you've crossed this threshold without a voltmeter and datasheet to tell you what the threshold is.