How to differentiate logs that don't have an ID to aggregate them separately on SQL?

I have put some logic as following, which may (or may not) help you to get min and max date. check demo here.

with cte as
(
  select
    *,
    dense_rank() over (partition by percentage order by rnk desc) as nrnk
  from
  (
    select
          *,
          (day - '2000-01-01'::date 
               - row_number() over (partition by percentage order by day)) as rnk
      from batteries
      where percentage is not null  
  ) t   
) 

select
  slot,
  min(day) as min_day,
  max(day) as max_day,
  percentage
from cte
where nrnk = 1    
group by
  slot, 
  percentage, 
  nrnk

Output:

| slot | min_day    | max_day    | percentage |
| ---- | ---------- | ---------- | ---------- |
| 0    | 2020-05-17 | 2020-05-17 | 1          |
| 0    | 2020-05-18 | 2020-05-18 | 2          |
| 0    | 2020-05-14 | 2020-05-14 | 95         |
| 0    | 2020-05-15 | 2020-05-15 | 96         |
| 0    | 2020-05-10 | 2020-05-12 | 97         |
| 1    | 2020-05-08 | 2020-05-10 | 10         |

If I understand the data correctly, you know there is a new battery when the value is NULL. If that is the indication, then you can calculate the sequence by calculating the number of NULL values up to each row (using a cumulative sum).

You have one additional step to get the last value and then aggregate:

select slot,
       min(day) filter (where percentage = last_percentage),
       max(day), last_percentage,
       sequence
from (select b.*,
             first_value(percentage) over (partition by slot, sequence order by day desc) as last_percentage
      from (select b.*,
                   count(*) filter (where percentage is null) over (partition by slot order by day) as sequence
            from batteries b
           ) b
       where percentage is not null
     ) b
group by slot, sequence, last_percentage
order by slot, sequence;

Here is a db<>fiddle.

In Redshift you just use case expressions or booleans:

select slot,
       min(case when percentage = last_percentage then day end),
       max(day), last_percentage,
       sequence
from (select b.*,
             first_value(percentage) over (partition by slot, sequence order by day desc) as last_percentage
      from (select b.*,
                   sum( (percentage is null)::int ) over (partition by slot order by day) as sequence
            from batteries b
           ) b
       where percentage is not null
     ) b
group by slot, sequence, last_percentage
order by slot, sequence;

This is an iterative process. If you want to do this in SQL, use a recursive query. In order to do so, number your rows per slot first, so as to get easy from one row to the next one

In the recursive part:

1. Compare a rows percentage with the one before and remember thus since when this percentage is current.
2. Whenever you hit a null percentage you get a new battery (increase the battery number).

At last: Show the row with the highest date per slot and battery.

with recursive numbered as
(
  select
    slot, day, percentage,
    row_number() over (partition by slot order by day) as rn
  from batteries
)
, cte(slot, battery, min_date, max_date, percentage, rn) as
(
  select slot, 1, day, day, percentage, rn
  from numbered
  where rn = 1
   union all
  select
    n.slot,
    cte.battery + case when n.percentage is null then 1 else 0 end,
    case when n.percentage = cte.percentage then cte.min_date else n.day end,
    n.day,
    n.percentage,
    n.rn
  from cte
  join numbered n on n.slot = cte.slot and n.rn = cte.rn + 1
)
select distinct on (slot, battery)
  slot, min_date, max_date, percentage, battery - 1 as sequence
from cte
order by slot, battery, max_date desc;

Result:

+------+------------+------------+------------+----------+
| slot |  min_date  |  max_date  | percentage | sequence |
+------+------------+------------+------------+----------+
|    0 | 2020-05-10 | 2020-05-12 |         97 |        0 |
|    0 | 2020-05-15 | 2020-05-15 |         96 |        1 |
|    0 | 2020-05-18 | 2020-05-18 |          2 |        2 |
|    1 | 2020-05-08 | 2020-05-10 |         10 |        0 |
+------+------------+------------+------------+----------+

Demo: https://dbfiddle.uk/?rdbms=postgres_12&fiddle=1ee3278d6f020f2d1a61bc93276965f5