Pulling the enable pin on Adafruit PowerBoost to low (It's connected to VBatt!)
Your biggest problem is that the module defaults to on. There is a 340kΩ resistor pulling the EN pin up to VBAT all the time, so to turn it off you have to actively override that pullup.
Ideally you would modify the module by removing that resistor (R13) and adding your own resistor between the ENABLE and GND pins. You can the override that with both a button and a P-channel FET in parallel to turn the module on.
For example:



simulate this circuit &ndash; Schematic created using CircuitLab
Now normally R1 pulls ENABLE to ground turning the module off.  You press SW1 and it pulls ENABLE up to VBAT and turns it on.  5V then gets supplied to the Arduino, at which point you need to set the GPIO to output and pull it HIGH.
This causes the gate of M2 to be pulled above its threshold voltage (make sure it has a threshold voltage between 0V and 3V) and it starts to conduct pulling the gate of M1 below its threshold voltage (must be between 0V and -2V) bypassing switch SW1 keeping the system enabled.
You can now let go of SW1 and it will all stay powered on until you either set the GPIO LOW or set it to an INPUT.
Note that resetting the Arduino (for programming, for example) would cause the boost module to shut down and require re-powering with SW1 (or through the GPIO if you have the Arduino powered through USB for programming of course).
However if you don't want to or can't modify the module then things get a little nastier.
To make it default to OFF R1 in the above schematic will have to be low enough value that the voltage presented at the ENABLE pin is below 20% of the battery voltage.  R1 and the existing 340kΩ pullup resistor form a voltage divider, so you need to calculate the size of resistor that will allow it to be below 20% VBAT.
Since a voltage divider just gives you a proportion of the input voltage whose ratio is the ratio of the &quot;top&quot; resistance in the divider to the total resistance we don't care what the voltage is, only the ratio.  And we don't really care about the exact value, only that the result is below 20%.  So we can pick a good common value.
However we don't want a value that is too low, because the voltage divider will be draining power from the battery all the time, so the higher the resistances the less the power drain will be.
For example if we pick the common value 47kΩ we see that:
    R2           47000
 ------- = --------------- = 0.11 (11%)
 R1 + R2    340000 + 47000

Which is below the 0.2 (20%) that we want, which is good.  Calculating the current at a typical battery voltage of 3.7V we get
     V          3.7
I = --- = -------------- = 0.000010 (10µA)
     R    340000 + 47000

10µA is not much, on the scale of things, but it's good to know about it when you're working from a battery since it will form part of your power budget calculations.