How to prevent `ls` from sorting the output?

With GNU ls, you could try the -U option:

-U: do not sort; list entries in directory order

(though here, we're not listing the content of directories, so the part that matters is do not sort).

$ ls -1dU /opt/foo/lib.jar /opt/bar/lib.jar
/opt/foo/lib.jar
/opt/bar/lib.jar

Slightly more portable (works with GNU and FreeBSD ls, but not with traditional ls implementations and is not POSIX either), you can use ls -1df:

$ ls -1df /opt/foo/lib.jar /opt/bar/lib.jar
/opt/foo/lib.jar
/opt/bar/lib.jar

You could create a wrapper function around ls that loops over the arguments (in the order they're given), calling ls on each:

myls() {
  for arg do
    ls -d -- "$arg"
  done
}

With a reference to Stéphane's answer in How to use arguments like $1 $2 ... in a for loop? for the for arg do syntax.

In zsh, to reduce a list of files to the ones that exist, you can use the (N) glob qualifier appended to each element of the list:

files=(/opt/foo/lib.jar /opt/bar/lib.jar)
that_exist=($^files(N))

(($#that_exist > 0 )) && print -rl -- $that_exist

Note that contrary to ls, it can tell whether a file exists or not even if you don't have search access to its parent directory (as long as you still have read access), as it looks for corresponding entries in the directory without trying to access the file.

See also:

that_exist_after_symlink_resolution=($^files(N-^@))

Which selects the files that are not (^) a symlink (@) after symlink resolution (-). Equivalent to using ls -Ld.

If you have a list of pathnames that you'd like to go through and test for existence, then I presume that you'd like to either do something fol the ones that do exist, for the ones that don't exist, or for both sets of pathnames.

Given a standard shell, this is best done in a loop.

for pathname in /the/first/path /the/second/path /some/other/path
do
    if [ -e "$pathname" ]; then
        printf 'Path "%s" exists\n' "$pathname"
    else
        printf 'Path "%s" does not exist\n' "$pathname"
    fi
done