Using the 0.1 Ohm shunt with the INA219 set to 40 mV full scale will work well.
Several of your statements are based on an incorrect understanding of the INA219 specification.
INA219 datasheet here
Acceptable full scale drop across the shunt can be adjusted to one of 320 mV, 160 mV, 80 mV or 40 mV (datasheet page 5). This voltage occurs due to voltage drop across the shunt when current flows.
Vshunt = I shunt x R shunt.
If the panel output is shorted via the shunt then loaded panel voltage will drop to whatever voltage is required to support the maximum available curremt - in that case, short circuit current Isc.
From supplied information I'll assume.
Isc ~~ 250 mA.
Voc ~= 6V.
Rshunt_available = Rsa = 0.1 Ohm.
At 250 mA and 0.1 Ohms the shunt voltage = V = I x R = 0.25A x 0.1 Ohm = 25 mV.
So even the most sensitive setting on the INA219 will accommodate Imax.
The 25 mV shunt drop at 250 mA will be similar at Imp as Imp is typically 80%-90% of Isc. So Vshunt % of operating voltage ~= 25 mV/4V x 100 = 0.6% of Voperating - so will have minimal affect on system operation so OK.
eg if using the panel to operate a resistive load at Vmp = 4V say then as Power = V^2/R, the power delivered with and without shunt will be in the ratio ((3.975)/4)^2 or 98.75%. So about 1.25% (or less) of available power will be lost in the shunt.
Small solar panels that chrage a battery are usually used with a series Schottky diode. If Vdiode ~= 0.3V the power loss in the diode is far greater than that in the shunt, so shunt power is essentially irrelevant.