一个选项是对第一查询返回的计数求和，然后将其与第二查询联合;另外，添加 constants，显示源代码：

select 'ab' what, (select count(*) from a) + (select count(*) from b) cnt from dual
union all
select 'xy', count(*) from xy;

您可以用途：

SELECT 'ab' AS type,
       COUNT(*) AS total
FROM   ( SELECT 1 FROM a UNION ALL
         SELECT 1 from b );
UNION ALL
SELECT 'xy', COUNT(*)
FROM   xy;

可以对三个联合Select语句的计数求和，并按源的组合对结果进行分组：

WITH 
    a AS
        ( Select LEVEL "A_ID", 'some column a' "COL_A" From Dual Connect By LEVEL <= 30 ),
    b AS 
        ( Select LEVEL "B_ID", 'some column b' "COL_B" From Dual Connect By LEVEL <= 20 ),
    xy AS 
        ( Select LEVEL "XY_ID", 'some column xy' "COL_XY" From Dual Connect By LEVEL <= 20 )

使用上述示例数据，如下所示：

SELECT
    CASE WHEN SOURCE IN('a', 'b') THEN 'ab' ELSE SOURCE END "SOURCE", 
    Sum(CNT) "CNT"
FROM
    (   Select 'a' "SOURCE", Count(*) "CNT" From a Union All
        Select 'b', Count(*) From b Union All
        Select 'xy', Count(*) From xy
    )
GROUP BY 
    CASE WHEN SOURCE IN('a', 'b') THEN 'ab' ELSE SOURCE END
--  
--  R e s u l t :
--  SOURCE        CNT
--  ------ ----------
--  ab             50 
--  xy             20

假设您的真实的查询可能要复杂得多，我认为我们不会试图改变它们，也不会以某种方式合并或拆分它们。
您的第一个查询返回两行。您希望得到它们的总和，因此必须聚合结果并使用SUM。
下面的查询对两个查询使用CTE（WITH子句中的子查询），然后对获取此总和的查询使用CTE。然后对最后的UNION ALL查询使用这些CTE。

with query1 (cnt) as (select count(*) from a union all select count(*) from b)
   , query2 (cnt) as (select count(*) from xy)
   , sumquery1 (total) as (select sum(cnt) from query1)
select 'ab' as tables, total from sumquery1
union all
select 'xy' as tables, cnt from query2
order by tables desc;