Charging Capacitor with Solar Panel

The answer is to use a charging chip that has the ability to back off on charging current when the input voltage from the solar panel begins to collapse.

While using a regular Li-Ion charging IC will technically "work" with a solar panel, it will take forever to charge in practice, and, as mkeith pointed out: "it will never get a chance to terminate charging".

The discussion about this on the Adafruit website is the best explanation of this phenomenon that I can recommend to others with this question:

Most people try to plug a solar panel directly into a lipo charger and while it sort of works, the battery takes forever to charge because the efficiency is terrible! That's because most lipo chargers are meant to plug into a USB port or wall, and are very simple in their design. USB ports supply 5V at up to 500mA and they're pretty solid - the voltage doesn't change much even at the max current draw. So when you plug a charger into a computer with a USB port, they just draw 500mA or so and happily chug away. Same goes for wall adapters. The voltage and current limits are kept steady. Solar panels are a little different, the voltage and current vary constantly depending on sunlight available. They are unstable! That instability confuses battery chargers, which causes them to do one of two things: rapidly turn on and off as they try to draw more current from the panel than possible and/or draw much less current than they can, to keep the voltage from collapsing

I was wondering what would happen if you took a 2.7v super-cap and began to charge it with 2.5V then while continuing to charge it the input voltage drops to 1V.

This question carries an assumption that has an error. Once the capacitor is charged to 2.5V, if you reduced the supply voltage, the capacitor would have charge taken away from it i.e. the action of lowering the terminal voltage "discharges" the capacitor.

Q = CV under all circumstances and assumign C is constant (not always a good assumption), if you lower the voltage you lower the charge stored in the device.

If you started with 1V and raised it to 2.5V then you are indeed charging the capacitor. Another formula derived from Q = CV is the formula for current into (or out of) a capacitor. Differentiate both sides with respect to time (assuming C is constant) and you get: -

\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

And of course dQ/dt = current. This tells you that if you apply a raming voltage (1 volt per second) a 1 farad capacitor, you will cause a constant current into the cap of 1 amp.