您可以使用相关子查询：

SELECT id1,
       id2,
       ( SELECT COUNT(*)
         FROM   table_name c
         WHERE  ( c.id1 = t.id1 AND c.id2 = t.id2 )
         OR     ( c.id1 = t.id1 AND c.id2 IS NULL )
         OR     ( c.id1 IS NULL AND c.id2 = t.id2 )
       ) AS cnt
FROM   table_name t
WHERE  id1 IS NOT NULL
AND    id2 IS NOT NULL;

其中，对于样本数据：

CREATE TABLE table_name (ID1, ID2, TIMESTAMP) AS
SELECT 'A',  1,     DATE '2023-09-01' + INTERVAL '11:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'A',  null,  DATE '2023-09-01' + INTERVAL '10:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'B',  null,  DATE '2023-09-01' + INTERVAL '09:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT null, 1,     DATE '2023-09-01' + INTERVAL '08:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'B',  2,     DATE '2023-09-01' + INTERVAL '07:00' HOUR TO MINUTE FROM DUAL;

输出：
| ID1| ID2| CNT|
| --|--|--|
| 一| 1 | 3 |
| B| 2 | 2 |
fiddle

您正在查找ID1和ID2都不为空的组。在这些组中，您希望计算与任一ID匹配的所有行。为此使用相关子查询。

select
  id1, id2,
  (
    select count(*)
    from mytable all_matches
    where all_matches.id1 = mytable.id1
       or all_matches.id2 = mytable.id2
  ) as total
from mytable
where id1 is not null and id2 is not null;

演示：https://dbfiddle.uk/UQcmFgGC

最后，我以下面的查询结束：

with subquery as (
    select * from (
    select COALESCE(id1, LAG(id1) OVER (PARTITION BY id2 order by id1)) as id1, 
           COALESCE(id2, LAG(id2) OVER (PARTITION BY id1 order by id2)) as id2,
    count(*) count 
    from T group by id1, id2 order by id1, id2
    )
)
select id1, id2, sum(count) from subquery group by id1, id2 order by id1, id2

感谢JonasMetzler的建议。

注意：这回答了5th revision of the question，而不是后来的编辑。
在Oracle 12中，您可以使用MATCH_RECOGNIZE进行逐行模式匹配：

SELECT *
FROM   table_name
MATCH_RECOGNIZE (
  ORDER BY timestamp
  MEASURES
    id1_id2.id1 AS id1,
    id1_id2.id2 AS id2,
    COUNT(*) AS cnt
  PATTERN ( (null_id1* id1_id2 null_id1*) | (null_id2* id1_id2 null_id2*) )
  DEFINE
    id1_id2  AS id1 IS NOT NULL AND id2 IS NOT NULL,
    null_id1 AS id1 IS NULL,
    null_id2 AS id2 IS NULL
)

其中，对于样本数据：

CREATE TABLE table_name (ID1, ID2, TIMESTAMP) AS
SELECT 'A',  1,     DATE '2023-09-01' + INTERVAL '11:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'A',  null,  DATE '2023-09-01' + INTERVAL '10:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'B',  null,  DATE '2023-09-01' + INTERVAL '09:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT null, 1,     DATE '2023-09-01' + INTERVAL '08:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'B',  2,     DATE '2023-09-01' + INTERVAL '07:00' HOUR TO MINUTE FROM DUAL;

输出：
| ID1| ID2| CNT|
| --|--|--|
| B| 2 | 2 |
| 一| 1 | 3 |
fiddle