Bash syntax error when "else" follows an empty "then" clause

It seems that you want to do a no-op if the file is open so you should add a :, which is a null command in bash:

if lsof "$filename" > /dev/null; then
  # file is open
  :
else
  printf 'deleting %s\n' "$filename"
  rm -- "$filename"
fi

If you don't use :, bash can not parse you code, and will show error like bash: syntax error near unexpected token 'else'.

Another alternative: reverse your logic.

if ! lsof "$filename" >/dev/null;then
    echo "deleting $filename"
    rm "$filename"
fi

Don't use command substitution on the output of find. Here, everything can be done with find:

find . -mtime 1 -type f ! -exec lsof -t {} \; -exec rm -f {} \; > /dev/null

With a few find implementations (including FreeBSD find where it comes from and GNU find), you can use -delete instead of -exec rm....

The reason you're getting an error is that there's no command between then and else and some shells (starting with the Bourne shell where that syntax comes from) require at least one (and a comment is not a command). Note that it is completely arbitrary and there is no reason why those shells would do that. yash and zsh don't have that limitation (if false; then else echo x; fi and even if false; then else fi work fine with them).

As others have said, you can use a noop command like : (or for nothing in; do nothing; done) or reverse the logic with the ! keyword (available in POSIX shells, but not the Bourne shell (you'll find that using : for that was common in that shell)). mksh and yash happen to support if false; then () else echo x; fi (I wouldn't rely on it as that could change in future versions though).

Another approach is with:

lsof... || {
  cmd1
  cmd2
}

though one difference is the overall exit status which will be that of lsof if lsof fails.