Как то так?
SQL> with main_table (id, sdate) as
2 (select 1, date '2018-01-02' from dual union all
3 select 2, date '2018-01-30' from dual union all
4 select 3, date '2018-07-25' from dual
5 ),
6 test_table (id, sdate) as
7 (select 1, date '2018-01-02' from dual union all
8 select 2, date '2018-08-30' from dual
9 )
10 select m.id,
11 m.sdate,
12 case when m.sdate = t.sdate then 'yes' else 'no' end status
13 from main_table m left join test_table t on t.id = m.id
14 order by m.id;
ID SDATE STATUS
---------- -------- ------
1 02.01.18 yes
2 30.01.18 no
3 25.07.18 no
SQL>
[ПРАВКА, после прочтения комментария - если вы найдете совпадение, вам вообще не нужен этот идентификатор]
Вот, пожалуйста:
SQL> with test (id, sdate) as
2 (select 1, date '2018-01-01' from dual union all
3 select 1, date '2018-01-02' from dual union all
4 select 1, date '2018-01-30' from dual union all
5 --
6 select 2, date '2018-10-01' from dual union all
7 select 2, date '2018-01-02' from dual union all
8 select 2, date '2018-01-30' from dual
9 )
10 select id, sdate
11 from test t
12 where not exists (select null
13 from test t1
14 where t1.id = t.id
15 and t1.sdate = to_date('&par_sdate', 'yyyy-mm-dd'));
Enter value for par_sdate: 2018-01-01
ID SDATE
---------- ----------
2 2018-01-30
2 2018-01-02
2 2018-10-01
SQL> /
Enter value for par_sdate: 2018-01-02
no rows selected
SQL>