Why doesn't source work when I call bash -c

If the goal is only to handle setting ENV before running a process and avoiding affecting the current shell then execing the bash executable might not be the most efficient way to go about it.

( . /dev/fd/4 && echo "$i" ) 4<<\SCRIPT
    i='i is set here in the subshell'
#END
SCRIPT

echo ${i?but i isnt set here because it was set in the subshell}

OUTPUT

i is set here in the subshell
sh: line 5: i: but i isnt set here because it was set in the subshell

You can of course substitute the link to the heredoc's file-descriptor with a regular file - I just used it to demonstrate it.

But if you do exec an outside process - such as bash or any other - rather than a shell builtin then you don't need the subshell.

one=1 two=2 \
    bash -c 'echo "${cmd=This is command #}$one" 
             echo "${cmd}$two"'
echo "${one?this var was only set for the execed ENV}"

OUTPUT

This is command #1
This is command #2
sh: line 2: one: this var was only set for the execed ENV

And if that outside process is bash or any other shell conforming to the POSIX standard of accepting -stdin as default, you can just write your script directly to a |pipe file...

{ echo PS1=
  echo 'echo "$PS1"'
  cat /etc/skel/.bashrc
  echo 'echo "$PS1"'
} | bash
echo "${PS1:?unset here again}"

OUTPUT

#blank line from first echo
[\u@\h \W]\$
sh: line 7: PS1: unset here again

As written, the shell is expanding $test before it is passed to your command. You can stop this from happening by using a ' instead of ".

bash -c 'source test.env && echo $test'