一个相当有效的解决方案是 exists 子查询：

select a.*
from a
where
    exists(select 1 from b where b.customer_id = a.customer_id and b.source = 'AA')
    and exists(select 1 from b where b.customer_id = a.customer_id and b.source = 'AB')
    and not exists(select 1 from b where b.customer_id = a.customer_id and b.source not in ('AA', 'AB'))

带索引的 b(customer_id, source) ，这应该运行得很快。
另一种选择是侵害：

select customer_id
from b
group by customer_id
having
    max(case when source = 'AA' then 1 else 0 end) = 1
    and max(case when source = 'AB' then 1 else 0 end) = 1
    and max(case when source not in ('AA', 'AB') then 1 else 0 end) = 0

这假设customer\u id/源组合没有重复项

select a.customer_id
  from A a join B b
    on a.customer_id = b.customer_id
 group by a.customer_id
-- both 'AA' and 'AB', but no other
having sum(case when b.source IN ('AA','AB') then 1 else -1 end) = 2

在联接之前进行聚合可能更有效：

select a.customer_id
from A a join 
  ( select customer_id
    from B b
    group by customer_id
    -- both 'AA' and 'AB', but no other
    having sum(case when source IN ('AA','AB') then 1 else -1 end) = 2
  ) b
on a.customer_id = b.customer_id

可以使用聚合：

select b.customer_id
from b
where b.source in ('AA', 'AB')
group by b.customer_id
having count(distinct b.source) = 2;

也就是说，你的版本应该可以用。然而，你应该学会使用适当的，明确的，标准的，可读的 JOIN 语法。但是，在这种情况下不需要连接。
如果只需要这两个源，则需要调整逻辑：

select b.customer_id
from b
group by b.customer_id
having sum(case when b.source = 'AA' then 1 else 0 end) > 0 and  -- has AA
       sum(case when b.source = 'AB' then 1 else 0 end) > 0 and  -- has AB
       count(distinct b.source) = 2;