For batteries or capacitors, you can draw a graph of the voltage against the amount of charge stored. For a 12v 100Ah lead-acid battery, the curve starts at maybe 10v, with virtually no charge stored, and then rises to just over 12v, with 360,000 coulombs stored. For the battery this curve is steep at both ends, but quite flat in the middle: and there are two curves, one for when you are putting energy in, and one for taking energy out, separated by about a volt-and-a-half.
With a capacitor, the "curve" is a straight line, starting at zero volts and zero coulombs, and in the case of a (huge!) 1F capacitor, reaching 12 coulombs at 12 volts.
Even if you could get a capacitor rated to store 360,000 coulombs at 12 volts (it would be a 30,000F capacitor - capacitors of a millionth of this size are still as big as a torch battery) the curves would meet at 12v, and again somewhere on the "flat" bit of the battery curve.
This matters because the total energy stored is equal to the area under the curve. (Don't worry about why - it's integral calculus.) When the battery is 50% discharged, the output voltage is maybe 11 volts, but the capacitor is only 6 volts. There is a lot of paper between the left hand side of the battery curve and the left hand side of the capacitor "curve". All that acreage of paper represents energy the battery can store, that the capacitor can't.
Another way to think of it is that the energy stored in the battery is approximately the average voltage times the total charge - call it 100Ah times 11v, or 1.1kWh. The same is true for the capacitor, but the average voltage is now 6v, and the energy only 0.6kWh.
To make things worse, several people here pour scorn on my notion of building an invertor that could deal with the extra voltage range produced by Edison batteries. Not without reason, either - power electronics is not an easy subject to dabble in, which is why most people buy their invertors, instead of building them. That problem would be nothing compared to the voltage range to get all the energy out of a capacitor. My 9-cell Edison battery's curve goes from 14.85v fully charged to 9v fully discharged, but your capacitor is at 0v fully discharged.
Getting that last bit of charge out would be a real struggle. You certainly couldn't do it with shop-bought invertors, and I think you'd really need to put your thinking head on to design something to do it efficiently.