You stated that the voltage source you have available is at least 5V, implying that it is perfectly capable of providing 2A without falling below that value.
The requirement that the load have at least 1.25A through it is fundamentally at odds with this above condition, since if the source is at least 5V, then you are able (with the right setup) to provide at least \$\frac{5V}{2.2\Omega-0.3\Omega} = 2.6A\$ under all conditions.
The actual voltage across the load, to sustain 2A through it, will be at a maximum with a load of greatest resistance, \$R_{LOAD(MAX)} = 2.2\Omega + 0.3\Omega = 2.5\Omega\$. The corresponding voltage will be:
$$ V_{LOAD} = I_{LOAD} \times R_{LOAD(MAX)} = 2A \times 2.5\Omega = 5V $$
Suddenly I see your problem. You cannot guarantee 2A through a 2.5Ω load using any voltage regulator with significant drop-out voltage. The question is, can a regulator maintain 1.25A through the load, even during drop-out?
The minimum voltage across the load to maintain 1.25A through it is:
$$ V = 1.25A \times 2.5\Omega = 3.13V $$
If your supply is guaranteed to be at least 5V, then you require a regulator with drop-out voltage at most:
$$ V_{DROPOUT} = 5V - 3.13V = 1.9V $$
There are plenty of commercial devices out there that easily beat this specification, but it I understand it would be better just to guarantee 2A at all times. Besides, a system design to deliberately operate a regulator in dropout is not ideal; I'm not sure that specifications for such operation are even available for regulators. It might be true, for some given regulator, that drop-out voltage is guaranteed to be remain constant as input falls below \$V_{OUT}+V_{DROPOUT}\$, but I won't make such a claim here.
Nominally, the voltage you need across the load is:
$$V = I\times R = 2A \times 2.2\Omega = 4.4V $$
Perhaps it's better to find a way to get exactly 4.4V across the load at all times. This is effectively asking for a regulated 4.4V output, even when the input source itself falls to 5V. That's a dropout of 0.6V. Regulators with such low dropout are available, but it might be difficult to find one that can promise that with a load current close to 2A.
You can find boost/buck regulators, though; DC-DC converters able to produce +5V given any input from, say, +3V and up. I will leave you to research that option.
The other way is to roll your own linear regulator, which is not difficult for resistive loads. The only reason a typical 3 or 4-terminal regulator has a "dropout voltage" in the first place is that the internal error amplifier must operate from whatever input potential is provided, and it is therefore unable to apply any potential greater than that to the base/gate of its own pass transistor.
If you are able to implement an error amplifier which operates from a supply with significantly greater potential than the required regulated output, then this problem goes away.
You can use any cheap, low power DC-DC converter to produce, say 12V (from whatever voltage your "5V or more" supply is producing) to operate the error amplifier, reference etc. You will be able to obtain a gate/base signal for the pass transistor(s) well beyond what a 3/4-terminal device could:
simulate this circuit – Schematic created using CircuitLab
This uses a closed loop to maintain a fixed voltage of 4.4V (produced by the TL431 reference IC) across the load, and therefore a fixed current through it of 2A, even as the input voltage (S) drops slightly below 5V. Here I sweep the input voltage from 0V to 12V, and plot voltage across, and current through the load:
As you can see, voltage remains exactly 4.4V for all inputs above 4.4V, only falling below that when the input supply drops below 4.4V. We essentially have a zero-volt dropout voltage regulator.
You never told us what the maximum source voltage will be. This is important, because it will determine the maximum power dissipation in M1. Here's how power in M1 varies with input voltage:
At 1W dissipation you should start thinking about heat-sinking. At 2W the heat-sink will have to be substantial. Higher than 4W is squarely in the realm of forced-air cooling (a fan).
You can mitigate heating by sharing the load across multiple pass transistors:
simulate this circuit
Resistors R2, R3 and R4 are necessary to balance current through each transistor. Without them, any mismatch in transistor parameters will cause one of them to dominate, and do all the heavy lifting, getting hot while the others sit back and relax.
