NiMH 1.2V 1/3 AA 280 mAh is it possible the claim made for size and capacity

Yes, you should be able to get a reasonable estimate of the battery's capacity as it is now - not what it was brand new.

It runs at 19 volts at 0.7 amps for 1/2 hour. So: 19 V x 0.7 amps x 0.5 hour = 6.65 Wh

6.65 Wh at 19 volts after it has been boosted from its 12 volt nominal state. You assume an 85% efficiency which is reasonable which means that you lost 15% in the boost, so 6.65/.85 = 7.82 Wh from the battery pack.

7.82 Wh for a 12 volt lithium 3s battery is really at 11.1 or 10.8 volts nominal. Since we don't know which one, and it doesn't make a big difference either way, I'll use 11.1 volts as the nominal voltage so: 7.82 Wh ÷ 11.1 v = 702 mAh.

Remember that it is a 3S2P so 702 ÷ 2 = 351 mAh per battery.

This seems a little low to me, so perhaps I have made a mistake somewhere, but I don't see where.

A way to confirm is to use the powerbank to charge a battery that you know the capacity of. If it charges that battery to 50% for example, you can then add the 15% loss for the boost to calculate the powerbank's capacity.

How about:

^{simulate this circuit – Schematic created using CircuitLab}

I have to say, I haven't tried it in the real-world with small cells yet, and it depends on the trickle level the battery will "eat" versus the very low Vf leakage in the diodes + leakage in the transistor.

But basically, when the battery is 2.5V the collector of Q1 will be up at 2.4V, the base will be held near 1.4V by the diodes, so R2 will "see 1V", which pulls 2mA through the battery, allowing sufficient base current into Q1 while still forcing the diodes into forward drop.

This will cause the transistor to go into voltage following emitter, which puts approximately 0.7V across R1, which turns that into approximately 7mA. Making a total of 9mA, which is sufficiently less than 1C to allow for "production errors". (Tweaking can be done by changing R1 and R2 and the supply voltage, of course).

Then, when the battery would be 4.3V, that leaves a collector voltage of about 0.6V, which should be in the region where the two diodes leak in the range of single μA, contributing only to a few tens of mV across R2, which then makes the transistor's base the main contributor there. At 0.55V base voltage that would be 100μA, that's possible with some transistors, but with this one not very likely. Looking at the Datasheet on page 4, figure 4, we can see that at 0.55V on the base the transistor can be considered off, and is not likely to conduct more that 50μA. I'd expect the total leakage to be in the 50μA range to be honest. But this would require checking, as I have only used this trick with 500mAh or higher, where 0.1mA is easy enough to handle if the battery can't.

It comes to a balance between low-operating voltage leakages and the chemical build quality of the cell. A decent Li-Ion battery can have leakages in the 2% per month range, which would be about (30 * 24 =) 720hours, which for your cell would mean:

2% of 13mAh = 0.26mAh
over 720 hours: 0.26mAh / 720hours =~ 0.36μA; which I would find a very amazing feat of engineering on their part in such a small cell, but it is possible.

In which case this would need a lot of tweaking, but with a tiny transistor and 0603 diodes and 0402 resistors it would be tiny at least. If you have cells to spare, why not try and see with a decent μA meter what the residual trickle is.

You can force the cell to plateau off sooner and more securely by adding one more resistor, theoretically at least:

^{simulate this circuit}

Where R3 would force the base to close, whenever the collector tries to leak when the battery gets close to 4.3V, in effect using the transistor's attempt at current amplification against it (although at 10μA collector currents amplification may be very, very low in this setup).

It also allows a higher voltage to begin with, as the extra resistor changes the balance a little there too.

If you haven't experimented enough yet, you can attempt the following for potentially lower leakages at 4.3V cell voltage:

^{simulate this circuit}

This is a bit of a freaky one, but the transistor will try to keep pulling enough current to keep the balance such that Vbase =~ Vemitter + 0.7V. The diodes will cause an upper limit, keeping the balance a little, even if the cell is in bad shape.

if the battery is at 4.2V, that leaves 0V across R1, which would again equate to base leakage, which will now be a little higher (as such you take 4.9V again).