Your "second" 'A' might better be denoted 'U'.

So you've found a basis for the column space of U (i.e., col(U)); they are columns 1, 3 and 4 of U.

But what you want is a basis for the column space of A (i.e., col(A)).

It's important to understand that A doesn't have the same column space as U.

The elimination process will leave some stuff unchanged (e.g., nullspace), but the columns will be entirely different.

Nevertheless, there's an important theorem telling us that when certain columns of U form a basis for col(U), the corresponding columns of A form a basis for col(A).