Another solution would be to get one or two large supercapacitors, for example let's say 25 F 2.7v- they're not that expensive these days, probably about 5-10$ each.
Put them in series to get 12.5 F 5.5v if your project needs more than 2.5v (you say battery but don't say what kind of battery) and charge them up to 5- 5.3v , or charge them to a bit above the level your project is designed for.
Then run your project for a few seconds and disconnect the project from the supercapacitors and measure the voltage on the supercapacitors.
You have original voltage, you have the voltage left , you can measure the energy consumed. Unlike batteries where the voltage will stay there for hours at low power consumption, the voltage drops pretty much linearly as energy is taken out from supercapacitors so you will notice a voltage difference right away (well, if your multimeter has a decent amount of counts, for example with my 22000 count uni-t ut61e this is a piece of cake).
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You can get a 0.1-0.33 ohm resistor to act as a current shunt and measure the voltage drop on it and if you want, you can put it on a scope and log the waveform. Note that there's voltage drop on the resistor, so that screws up the estimation a bit, especially with 5ohm as you did. The more power consumption, the more voltage drop on the resistor (Ex for 5 ohm, for 100mA you have 0.5v drop, so if your battery is 3v you now suddenly have 2.5v instead of 3v)
For example, if your project uses a switching regulator to convert battery voltage to some particular voltage, then just the change in inpuut voltage will screw up the efficiency of your regulator, which will make you estimate the battery life incorrectly.