Вам не хватает псевдонима для подзапроса в скобках и предложение on
:
SELECT
c.TABLE_NAME,
c.COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS c
INNER JOIN
(SELECT
COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
GROUP BY COLUMN_NAME
HAVING COUNT(*) = 1
) x -- Alias added here
ON x.COLUMN_NAME = c.COLUMN_NAME -- ON caluse