Why does current not increase when batteries are connected in parallel?
First off, I want to warn you just a little bit about putting battery systems in parallel. It's usually not a good idea because often the two batteries (or battery systems) don't have exactly the same voltage. If they are different, then the one with the larger voltage will supply some current into the battery with the lower voltage and this often isn't a good thing. It also messes up your experiment, somewhat, because it adds another complication to it.

In this case, you are curious and imagine that two batteries in parallel can supply more current. So it doesn't serve your purposes to use only one in your experiment because it doesn't test your assumptions. So you have to do it the way you did. But I just want you to also realize that there is another unknown (to you) factor that you aren't accounting for in your experimental design. But it's not enough to worry about, for now.

So set that aside...

Let me suggest a new idea for you to consider. Suppose that the green LED you have requires exactly \$1.9\:\text{V}\$ in order to "turn on" and that it also has, unknown to you, an internal resistor of exactly \$50\:\Omega\$. You can't get inside the LED to see these things. But let's say, as a thought experiment, that this is the way this particular LED works.

Also, let's assume that your battery systems provide exactly \$2.9\:\text{V}\$ each. If you put them in series then you are applying \$5.8\:\text{V}\$ to the circuit. If you put them in parallel then you are applying \$2.9\:\text{V}\$ to the circuit. The only difference here may be the current compliance (the ability to supply more or less current to a load, if required.)

Your assumption is that if the current compliance is more, then the current is more. But that may be true sometimes and not others. So, for now, let's use my above idea about the LED and see where that takes us.

Your series circuit includes also a \$100\:\Omega\$ resistor. Taken together with my hypothetical internal \$50\:\Omega\$ resistor inside the LED, there is a total series resistance in the circuit of \$150\:\Omega\$ (let's just assume I'm right for now.) Plus, the LED itself (the one inside the package that you cannot actually touch) also requires (subtracts) \$1.9\:\text{V}\$ from the applied voltage before we can compute the current. (You can see that the LED is ON in both cases, so this must be true if my assertion is correct.)

So in the batteries-in-parallel case you have \$I_\text{parallel}=\frac{2.9\:\text{V}-1.9\:\text{V}}{150\:\Omega}\approx 6.7\:\text{mA}\$ and in the batteries-in-series case you have \$I_\text{series}=\frac{2\cdot 2.9\:\text{V}-1.9\:\text{V}}{150\:\Omega}\approx 26\:\text{mA}\$.

This would appear to predict your measurements within a reasonably small error.

So which idea do you think works better here? Your thoughts about two parallel battery systems doubling the current? Or my suggestion about how an LED may behave? Do you have still further ideas that you may want to consider? How might you test or validate my above suggestion? Can you think of another way to change your circuit that might put my suggestion to another test to see if it still holds up? Or can you think of another voltage measurement or current measurement you might try to test it?