Finding missing dates in a sequence

To get a simple list of the IDs with gaps, with no further details, you need to look at each ID separately, and as @mikey suggested you can count the number of months and look at the first and last date to see if how many months that spans.

If your table has a column called month (since date isn't allowed unless it's a quoted identifier) you could start with:

select id, count(month), min(month), max(month),
  months_between(max(month), min(month)) + 1 as diff
from your_table
group by id
order by id;

        ID COUNT(MONTH) MIN(MONTH) MAX(MONTH)       DIFF
---------- ------------ ---------- ---------- ----------
       123            8 01-JUN-14  01-JUL-15          14
       456            7 01-MAR-14  01-NOV-14           9
       789            7 01-MAR-14  01-SEP-14           7

Then compare the count with the month span, in a having clause:

select id
from your_table
group by id
having count(month) != months_between(max(month), min(month)) + 1
order by id;

        ID
----------
       123
       456

If you can actually have multiple records in a month for an ID, and/or the date recorded might not be the start of the month, you can do a bit more work to normalise the dates:

select id,
  count(distinct trunc(month, 'MM')),
  min(trunc(month, 'MM')),
  max(trunc(month, 'MM')),
  months_between(max(trunc(month, 'MM')), min(trunc(month, 'MM'))) + 1 as diff
from your_table
group by id
order by id;

select id
from your_table
group by id
having count(distinct trunc(month, 'MM')) !=
  months_between(max(trunc(month, 'MM')), min(trunc(month, 'MM'))) + 1
order by id;

You could use a Lag() function to see if records have been skipped for a particular date or not.Lag() basically helps in comparing the data in current row with previous row. So if we order by DATE, we could easily compare and find any gaps.

select * from 
   (
    select ID,DATE_, case when DATE_DIFF>1 then 1  else 0 end comparison from
        (
          select ID, DATE_ ,DATE_-LAG(DATE_, 1) OVER (PARTITION BY ID ORDER BY DATE_)  date_diff from trial
        )
    )
    where comparison=1 order by ID,DATE_;

This groups all the entries by id, and then arranges the records by date. If a customer is always present, there would not be a gap in his date. So anyone who has a date difference greater than 1 had a gap. You could tweak this as per your requirement.

EDIT : Just observed that you are storing data in mm/dd/yyyy format, when I closely observed above answers.You are storing only first date of every month. So, the above query can be tweaked as :

select * from 
   (
    select ID,DATE_,PREV_DATE,last_day(PREV_DATE)+1 ABSENT_DATE, case when DATE_DIFF>31 then 1  else 0 end comparison from
        (
          select ID, DATE_ ,LAG(DATE_,1)  OVER (PARTITION BY ID ORDER BY DATE_)  PREV_DATE,DATE_-LAG(DATE_, 1) OVER (PARTITION BY ID ORDER BY DATE_)  date_diff from trial
        )
    )
    where comparison=1 order by ID,DATE_;

Oracle Setup:

CREATE TABLE your_table ( ID, "DATE" ) AS
SELECT 123, DATE '2015-07-01' FROM DUAL UNION ALL
SELECT 123, DATE '2015-06-01' FROM DUAL UNION ALL
SELECT 123, DATE '2015-05-01' FROM DUAL UNION ALL
SELECT 123, DATE '2015-04-01' FROM DUAL UNION ALL
SELECT 123, DATE '2014-09-01' FROM DUAL UNION ALL
SELECT 123, DATE '2014-08-01' FROM DUAL UNION ALL
SELECT 123, DATE '2014-07-01' FROM DUAL UNION ALL
SELECT 123, DATE '2014-06-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-11-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-10-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-09-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-08-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-05-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-04-01' FROM DUAL UNION ALL
SELECT 456, DATE '2014-03-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-09-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-08-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-07-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-06-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-05-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-04-01' FROM DUAL UNION ALL
SELECT 789, DATE '2014-03-01' FROM DUAL;

Query:

SELECT ID,
       MIN( missing_date )
FROM   (
  SELECT ID,
         CASE WHEN LEAD( "DATE" ) OVER ( PARTITION BY ID ORDER BY "DATE" )
                     = ADD_MONTHS( "DATE", 1 ) THEN NULL
              WHEN LEAD( "DATE" ) OVER ( PARTITION BY ID ORDER BY "DATE" )
                     IS NULL THEN NULL
              ELSE ADD_MONTHS( "DATE", 1 )
              END AS missing_date
  FROM   your_table
)
GROUP BY ID
HAVING COUNT( missing_date ) > 0;

Output:

        ID MIN(MISSING_DATE) 
---------- -------------------
       123 2014-10-01 00:00:00 
       456 2014-06-01 00:00:00 

    with 
    -- Unique dates in the source table
    unique_dates as(
      SELECT 
      distinct event_date
      FROM date_table
    ),
    -- Find the next available date for each date
    dates_lagged as (
      SELECT *
      , LAG(event_date,1) OVER (ORDER BY event_date DESC) as next_day
      FROM unique_dates
    )
    -- Gives you the beginning of each missing date timeframe and the number of days missing
    SELECT event_date
    , next_day
    , date_diff(next_day,event_date,DAY) as missing_days 
    FROM dates_lagged
    WHERE date_diff(next_day,event_date,DAY)!=1