The issue you're confused about seems to be the difference between power and energy.
Energy is how much work you can do. Common units are joules or watt-hours.
Power is how fast you do work. It's a rate of change. Common units are watts or horsepower. Horsepower is probably an instructive unit to consider. Say you wanted to move a large pile of straw. Whether it's moved by a horse or a housecat doesn't affect the amount of work done. But the horse does it faster, because it's a more powerful animal.
For the purposes of discussing grid electricity consumption, watts (W) and kilowatt-hours (kWh) are the most common units used. To know how much energy is consumed, multiply the power by the time. 100 W x 1 hour is 100 watt-hours, or .1 kWh. In short, the relationship between power and consumption is time.
A 100W bulb consumes 100W assuming the voltage across it is what's specified on the package, which is usually 120V in my experience. If the voltage at your socket is lower, the bulb will consume less power. It's approximately a fixed resistance, so the power consumed is
$$
P=\frac{V^2}{R}
$$
As an aside, remember energy conservation. If something consumes 100W, that energy is being converted to some other form. Either it gets stored (potential energy), it's used (light, motion, etc.), or it's wasted as heat. For an incandescent bulb, ~90% of the power consumed is converted to heat. So a 100W incandescent bulb consumes 100W, but only outputs 10W of light. It gets hot because the other 90W is being wasted. Which is why CFL's run so much cooler and consume less power for the same light output.
Sounds like he has micro inverters. Micro inverters directly attach to each panel and are wired in parallel. They simply turn DC into AC and feed it to the homes electrical system and to the grid depending on load. The smart meter tracks how much power is consumed from the grid vs fed back to the grid (excess). They also do not work in a blackout, they must see grid power to turn on.
How about storing that excess energy in a battery bank? The higher the excess, the more charge current is delivered to the battery (via PWM switching). The picaxe can monitor the excess and control a charge circuit to compensate for the load. Then you can use an off the shelf inverter to power appliances from the battery bank. The benefit of this system is that not only is it simple, but also gives you battery backup for blackout protection. Your friend can also use relays to switch certain appliances between battery and grid depending on the time of day. So during the day excess is dumped to batteries. At night when there is no more excess, relays can switch certain appliances or lighting circuits to battery and when the batteries run low, switch back to grid. Examples of relayed appliances could be: refrigerator, air conditioner, well water pump (if he has), lights, washing machine etc. The same grid->battery charger can also detect cloudy or low PV producing days and keep the battery bank charged in case of a blackout.
The batteries can get expensive but its a simple way to catch the excess and you have blackout protection.
Best Answer
Suppose that the declared power is a nominal voltage of 12V, and the series circuit is powered by a 12 V battery.
In this case, each lamp can be modeled with a resistor:
$$ R_{L1} = \dfrac{12^2}{100} = 1.44\,\Omega $$
$$ R_{L2} = \dfrac{12^2}{200} = 720\,\mathrm{m}\Omega $$
That is, when we apply to each lamp 12 V, the power consumption is the nominal (100 W and 200 W), and this power corresponds to the calculated resistance values.
For simplicity, we assume that the resistance is linear (although in reality depends on factors such as the temperature of the filament). Connecting these two resistors in series, we obtain a total resistance:
$$ R_{eq} = R_{L1} + R_{L2} = 2.16\,\Omega $$
so that the circuit current flows as
$$ I = \dfrac{12}{R_{eq}} = 5.556 \,\mathrm{A} $$
This current, dissipates power in each resistor as:
$$ P_{L1} = I^2\cdot R_{L1} = 44.44\,\mathrm{W}\\ P_{L2} = I^2\cdot R_{L2} = 22.22\,\mathrm{W} $$
As you can see, on each lamp, less than the rated power is developed, which results in lower brightness. It is important to remember that these results are only approximate because a lamp is not a completely linear element.
As additional information, we can obtain the necessary current to each lamp at rated:
$$ I_{L1nom} = \dfrac{100}{12} = 8.333\,\mathrm{A}\\ I_{L2nom} = \dfrac{200}{12} = 16.67\,\mathrm{A}\\ $$
which are much higher than those estimated for the series connection of two lamps values.
Remember, this is only one example at a given supply voltage, but can serve as a guide for your analysis.