This circuit uses a FET to drive a transistor, but you can replace the transistor with a darlington, so that you get kind of an hybrid superdarlington.
This should solve the opamp output current limitation.
All of these solutions will waste exactly the same amount of electrical energy, and generate the same amount of heat. Each of your circuits just changes which component gets hot.
These are all what's called a linear current source. A linear current source works (by definition) by converting excess voltage to heat. If your load requires 2V to reach the desired current of 5A, and the supply is 5V, the heat (power) \$P\$ will be the excess voltage \$E\$ times the current \$I\$:
\$ P = IE = 5A(5V-2V) = 5A\cdot 3V = 15 W\$
No way around that with any linear current source. You can spread it out or move it around different components, but you will never reduce it. Blame physics. The energy has to go somewhere.
If you want to reduce the wasted energy, you probably want a switched mode power supply (SMPS). The design of such is worthy of an entire book, but if you want a quick introduction, I suggest you read How can I efficiently drive an LED? Although you aren't driving an LED, the problem is essentially the same, since LEDs are ideally also driven with a current source.
However, since it sounds like your load is a fixed \$1 \Omega\$ resistor, you don't really need a current source. A voltage source would do just as fine, since a resistor is a current - voltage converter, by Ohm's law:
\$ E = IR \$
If it's allowable in your application, a simpler solution than an SMPS is to just switch the full battery voltage over your load on and off rapidly. If your supply is 5V then this will deliver 5A to your load. You can deliver 5A or 0A with low losses, and if you need something between those, then you switch it on and off rapidly, so the average current is your desired value. For many applications, this is good enough. If not, a SMPS is basically that, with an inductor added to smooth the current out to the average value across switching cycles.
Best Answer
The lazy approach would be: -
You can find all their integrated current sources here