此查询，使用分析 lead() 做这个工作。列 note 显示行是来自您的数据还是缺少间距：

select id, d1, d2, case dir when 3 then 'GAP' end note 
  from (
    select id, 
           case when dir = 2 
                 and lead(dir) over (partition by id order by dt) = 1
                 and lead(dt) over (partition by id order by dt) <> dt + 1
                then dt + 1 
                else dt
           end d1,
           case when dir = 2 
                 and lead(dir) over (partition by id order by dt) = 1
                 and lead(dt) over (partition by id order by dt) <> dt + 1
                then 3 
                else dir
           end dir,
           case when lead(dir) over (partition by id order by dt) = 1 
                then lead(dt)  over (partition by id order by dt) - 1
                else lead(dt)  over (partition by id order by dt) 
            end d2
      from (
        select * from a unpivot (dt for dir in (validfrom as 1, validto as 2)) union 
        select * from b unpivot (dt for dir in (validfrom as 1, validto as 2)) ) )
  where dir in (1, 3)

dbfiddle演示
一开始数据是不固定的，只是为了把所有的日期都放在一列中，这样更便于进一步分析。并集删除重复的值。列 dir 如果这是 from 或者 to 日期。那么 lead 根据该方向的类型，应用逻辑。我觉得可以简化一下：）