Usando SUBSTRING_INDEX :
UPDATE table1
SET column1 = REPLACE(
column1,
SUBSTRING_INDEX(column1, '/', 2),
'newsite.com'
)
WHERE column1 LIKE 'example.com/%/'
Isso deve respeitar sua estrutura de subpastas.
UPDATE table1
SET column1 = REPLACE(
column1,
SUBSTRING_INDEX(column1, '/', 2),
'newsite.com'
)
WHERE column1 LIKE 'example.com/%/'