Why can’t I simulate a current sensing IC (LT6100) using Proteus 8

I haven't worked with Proteus but most electronic CAD packages have similar behavior when dealing with individual gates.

That is: each gate is individual. You place all of the individual gates in your schematic as required. Then you gather the gates into the desired packages.

My CAD package names this process as "packaging". You do this in the stage between completing the schematic diagram and starting to PCB layout.

Packaging assigns the individual gates to the physical chip.

The information about this process should be in the user manual for your particular CAD program.

This is mostly a question of:

1. How much voltage are you willing to drop?
2. How much voltage do you need to drop?
3. How much power can you afford to lose?

If you're measuring 10 A, you're going to dissipate 100 watts through your resistor, but for you, 1 Ω may be perfectly fine. Let's run the numbers:

1. At 250 mA, the voltage drop is 250 mV. From what I understand, this is the highest current you are interested in. Since you seem to have control over your input voltage, this doesn't sound like a problem.
2. You have done this calculation already. At 250 mV and a gain of 20, you're looking at 0 to 5 volts full range.
3. The power through the resistor is I²R, 62.5 mW. As long as you use a large enough resistor (physically), this is not a problem. Even a tiny 0805 is usually rated at 125 mW, but you may want a larger package to combat self-heating.

I'm afraid that another stage will add noise, offset error and gain error.

Sure, it might. But the AD8418 isn't perfect either. The key to a good design is to calculate what kind of accuracy you need. The offset and gain error can be ignored: That's just a multiplication and addition in the microcontroller. The extra noise, well. That can be filtered out, if it's a problem.

Should I go ahead and use a "stable" 1 Ohm resistor or try another, more complex, solution?

It seems to me that using 1 Ω is perfectly reasonable.