LDO selection based on power dissipation
Your reasoning is fine, and the calculations are correct. You're perfectly right. And 1.1W can be dissipated by a SOT-223 package, provided that you have sufficient copper area under the package. There are some resources on the internet giving hints on the minimum copper area you need for a given power with this package:


TI AN-1028
mbedded.ninja


The only thing that needs clarification is that it's not false advertising. MIC5225 can handle 16V input. It can also handle 150mA. The thing is: you can't have both at the same time with this package, indeed. Either you use it with high input voltage but low current (some applications just need a few tens of mA), or, if you need higher current, you need to have lower input voltage.

Note that you can probably have both conditions true for a very short time, however.

A trick that you can use with linear regulators in such a situation is to use a low value, high-power resistor in series with the input. This resistor will dissipate most of the heat when the current raises. You just have to calculate its value so that the voltage drop across this resistor, when the current is at its maximum value, is lower than the difference between your input voltage and the minimum required voltage at the regulator input (use ohm's law). The only drawback would be slightly worse load regulation.