How do I grep without leading whitespaces?

You can just eliminate them using sed

grep blah filename.foo | sed -e 's/^[ \t]*//'

That will remove the leading whitespaces from the output

Assuming you're looking for pattern re (a basic regular expression) in one file, and you'd like to strip leading whitespace from all matching lines:

sed -n -e 's/^[[:blank:]]*//' -e '/re/p' thefile.c

(actually, this strips all leading whitespaces first, and then looks for the pattern, but the result is the same)

To post-process the grep output instead (as in your edited question):

grep -e 're' -- * | sed 's/:[[:blank:]]*/: /'

The pattern [[:blank:]]* matches zero or more spaces or tabs.

If you insert a tab instead of a space after the :, you additionally get some of the nice even indentation you requested.

Create test files

echo -e "\t   foo-somethingfoo" >something.foo
echo "    bar-bar-somethingbar" >something.bar_bar
echo "baz-baz-baz-somethingbaz" >something.baz_baz_baz
echo "  spaces    something  s" >something.spaces

produce full glorious colour :)

grep --colour=always "something" something.* | 
 sed -re  's/^([^:]+):(\x1b\[m\x1b\[K)[[:space:]]*(.*)/\1\x01\2\3/' |
   column -s $'\x01' -t

output (run it to get the colour).

something.bar_bar      bar-bar-somethingbar
something.baz_baz_baz  baz-baz-baz-somethingbaz
something.foo          foo-somethingfoo
something.spaces       spaces    something  s

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Tested in gnome-terminal, konsole, terminator, xterm