Such a control circuit is possible, but a better idea is to leave the panels fixed in one configuration and deal with the resulting change in voltage with the right kind of switching power supply. If you want to get fancy, you could even implement maximum power point tracking, or something reasonably close.
The switching power supply can't make more power than what the panels produce, but it can convert (with a little loss) from whatever voltage and current the panels want to produce to a different controlled voltage or current (with the other limited by the available power).
If the series configuration always produces more voltage than you need even in low light, then that's how you should configure the panels. The switching power supply then always makes a lower voltage, which keeps it simple. That is called a "buck" regulator, with much information about that out there.
It would be helpful to say what voltage you ultimately want, and what current range is useful at that voltage.
Added:
You now say this is to power a 12V system, which presumably runs from lead-acid batteries although you didn't say that.
One useful feature of lead acid batteries is that they can take reasonable charge current even when full and still regulate the voltage well enough for most purposes. Given that, a really simple solution is to wire the solar panels in series to get the higher voltage you talked about, which is always a bit more than the 12V battery level even on cloudy days (when light is really low the voltage will be lower, but then there is so little power to be irrelevant), and just connect this to the 12V rail with a Schottky diode. That will not use the panels most efficiently in high illumination, but probably not so bad on cloudy days from the numbers you give.
A buck converter that runs the panels at the best efficiency for the given insolation and then dumps whatever current it can onto the 12V rail should be a bit more efficient. With a decent converter design, the extra loss in the switcher should be more than offset by running the panels at their optimum efficiency.
However, if you have more than enough power in full sunlight and the real problem is when it's cloudy, maybe the dumb series connection (with a Schottky diode to prevent reverse current when dark) will do it. I'd probably be tempted to try that first and see what you get and how efficient the whole system is on cloudy days when it really matters.
Your post deals with several different aspects, which I'll clarify:
1) The power necessary to charge the battery can be almost anything. You can charge a battery with 1W or with 5W. The voltage must always be the same. If you use less watts, the resulting lower current means less charge and therefore longer chargning time. There is a maximum current at which a battery can be charged. Ususally this is 1C, where the C is the capacity of the battery which is 1650mAh. So, battery max current is 1650mA. In reality, it can be C/2 or other values depending on the battery chemistry. Chargers must remain below this
2) CNET must be wrong about 55W. At 3.7V you're talking about 14.86A which is very high and a battery of 1650mAh would likely get destroyed. I've worked with Cell Phones and although the peaks are high, they are never this high. Given a 1650mAh battery, you can draw 1650mA in one hour, or double in half an hour. In reality, the rate of discharge ends up changing the nominal capacity, so this isn't linear but an approximation.
As you calculated, the wattage is about 0.76W at 3G.
Best Answer
Your understanding of Watts is correct. Power = Voltage * Current. Power is measured in watts, so your "xy" analogy is accurate.
One thing that you should keep in mind is losses that occur between energy conversions. Even if a device is rated at Z watts, it may sometimes draw more than Z (or sometimes less than Z). Try to have a little extra margin on your solar array output wattage and you will be fine. Happy Travels!