不能使用JSON函数拆分字符串。
您可以使用简单的字符串函数（比正则表达式快得多）和递归子查询分解子句拆分字符串：

WITH rsqfc (scp, aud, spos, epos) AS (
  SELECT scp,
         aud,
         1,
         INSTR(scp, ' ', 1)
  FROM   JSON_TABLE(
           '{"s": "s1 s2 s3","a": ["a1","a2","a3"]}' FORMAT JSON,
           '$[*]'
           COLUMNS(
             scp VARCHAR2 PATH '$.s',
             NESTED PATH '$.a[*]' COLUMNS(
               aud VARCHAR2 PATH '$[*]'
             )
           )
         )
UNION ALL
  SELECT scp,
         aud,
         epos + 1,
         INSTR(scp, ' ', epos + 1)
  FROM   rsqfc
  WHERE  epos > 0
)
SELECT CASE epos
       WHEN 0
       THEN SUBSTR(scp, spos)
       ELSE SUBSTR(scp, spos, epos - spos)
       END AS scp,
       aud
FROM   rsqfc;

在Oracle 12中，您还可以使用CROSS APPLY：

SELECT REGEXP_SUBSTR(scp, '[^ ]+', 1, lvl) AS scp,
       aud
FROM   JSON_TABLE(
         '{"s": "s1 s2 s3","a": ["a1","a2","a3"]}' FORMAT JSON,
         '$[*]'
         COLUMNS(
           scp VARCHAR2 PATH '$.s',
           NESTED PATH '$.a[*]' COLUMNS(
             aud VARCHAR2 PATH '$[*]'
           )
         )
       )
       CROSS APPLY (
         SELECT LEVEL AS lvl
         FROM   DUAL
         CONNECT BY LEVEL <= REGEXP_COUNT(scp, '[^ ]+')
       );

两者输出：
| SCP| AUD|
| --|--|
| S1| A1|
| S1| A2|
| S1| A3|
| S2| A1|
| S2| A2|
| S2| A3|
| S3| A1|
| S3| A2|
| S3| A3|
fiddle