MySQL 8.0

By using db<>fiddle, you agree to license everything you submit by Creative Commons CC0.

CREATE TABLE external_user (id INTEGER PRIMARY KEY, email VARCHAR(100))
INSERT INTO external_user VALUES
  (1, 'email@domain1.com'), (2, 'email@domain1.com'), (3, 'email@domain1.com'), 
  (4, 'email@domain2.com'), (5, 'email@domain2.com');

Records: 5  Duplicates: 0  Warnings: 0


SELECT email, 
       SUBSTRING_INDEX(GROUP_CONCAT(id ORDER BY id DESC), ',', -COUNT(*) + 1) AS ids 
FROM external_user 
GROUP BY email 
HAVING COUNT(*) > 1;


email
ids


email@domain1.com
2,1

email@domain2.com
4