Although I haven't done it, I would think that a small N-channel FET and a resistor in parallel with the battery would work. Measure the voltage of the battery with no (or a known) load. Turn on the FET momentarily to increase the load and measure the voltage. Turn off the FET and do the calculations. It is Ohm's law at that point. If the FET added 1A of current draw, and the "loaded" voltage was 1V less than the unloaded voltage, then the ESR (impedance) of the battery is 1 Ohm. All the formulas you need are here-

http://en.wikipedia.org/wiki/Voltage_divider