You're on the right track. The catch is that your test of "does the following directory NOT exist in this target" can't be expressed within find's conditions in such a way as to return only the top-level directory. So you need to nest, one way or another.

One strategy would be to use a for loop in bash:

$ mkdir foo bar baz one two
$ mkdir bar/bin baz/bin
$ for d in /home/*/; do find "$d" -type d -name bin | grep -q . || echo "$d"; done
foo/
one/
two/

This uses pathname expansion (globbing) to generate the list of directories to test, and then checks for the existence of "bin". If that check fails (i.e. find outputs nothing), the directory is printed. Note the trailing slash on /home/*/, which ensures that you will only be searching within directories, rather than files that might accidentally exist in /home/.

Another possibility might be to use nested finds, if you don't want to depend on bash:

$ find /home/ -type d -depth 1 -not -exec sh -c "find {}/ -type d -name bin -print | grep -q . " \; -print
/home/foo
/home/one
/home/two

This roughly duplicates the effect of the bash for loop above, but by nesting find within find -exec. It uses grep -q . to convert the output of find into an exit status that can be used as a condition for the outer find.

Note that since you're looking for a bin directory, we want to use test -d rather than test -e (which would also check for a bin file, which probably does not matter to you.)

Another option is to use bash process redirection. On multiple lines for easier reading:

cd /home/
comm -3 \
  <(printf '%s\n' */ | sed 's|/.*||' | sort) \
  <(find */ -type d -name bin | cut -d/ -f1 | uniq)

This unfortunately requires you to change to the /home directory before running, because of the way it strips off subdirectories. You can of course collapse this into a big long one-liner if you feel so inclined.

This comm solution also has the risk of failing on directories with special characters in their names, like newlines.

One last option is bash-only but more than a one-liner. It involves subtracting the directories containing "bin" from the full list. It uses an associative array and globstar, so it depends on bash version 4.

#!/usr/bin/env bash

shopt -s globstar

# Go to our root
cd /home

# Declare an associative array
declare -A dirs=()

# Populate the array with our "full" list of home directories
for d in */; do dirs[${d%/}]=""; done

# Remove directories that contain a "bin" somewhere inside 'em
for d in **/bin; do unset dirs[${d%%/*}]; done

# Print the result in reproducible form
declare -p dirs

# Or print the result just as a list of words.
printf '%s\n' "${!dirs[@]}"

Note that we're storing directories in the array index, which (1) makes it easy for us to find and delete items, and (2) insures unique entries, even if one user has multiple "bin" directories under their home.

cd /home
find . -maxdepth 1 -type d ! -name . | sort > a
find . -type d -name bin | cut -d/ -f1,2 | sort > b
comm -23 a b

Here, I'm making two sorted lists. The first contains all the home directories, and the second contains the top parent of any bin subdirectory. Finally I output any items from the first list not present in the second.