Wan to get group by date and status from the data

You can convert your 'created_at' column to datetime using pd.to_datetime, filter for your 'success' rows, and then use groupby per month with the aggregation of count:

df['created_at'] = pd.to_datetime(df['created_at'])
out = df.loc[df.status.eq('success')].groupby(df.created_at.dt.month).status.count()

prints:

   created_at  success
0           1        2
1          12        1

# as a dict

>>> out.set_index('created_at').to_dict()
{'success': {1: 2, 12: 1}}

Then you can just convert the month numbers to month names as shown here

You have to parse the date using datetime.strptime first, and use itetools.groupby to group based on month.

>>> from datetime import datetime
>>> import calendar
>>>
>>> from itertools import groupby
>>>
>>> data = [
...     {"name": "aa", "created_at": "2022-01-17 07:38:26.403Z", "status": "success"},
...     {"name": "bb", "created_at": "2021-12-1 07:38:26.403Z", "status": "failed"},
...     {"name": "kk", "created_at": "2022-01-13 07:38:26.403Z", "status": "success"},
...     {"name": "ll", "created_at": "2021-12-17 07:38:26.403Z", "status": "success"},
... ]
>>>
>>>
>>> def custom_key_fun(row):
...     created_at_as_datetime = datetime.strptime(
...         row["created_at"], "%Y-%m-%d %H:%M:%S.%fZ"
...     )
...     month = calendar.month_abbr[created_at_as_datetime.month]
...     return month
...
>>>
>>> result = {
...     key: sum(1 for row in group if row["status"] == "success")
...     for key, group in groupby(sorted(data, key=custom_key_fun), key=custom_key_fun)
... }
>>>
>>> print(result)
{'Dec': 1, 'Jan': 2}

Load Pandas

import pandas as pd
from datetime import datetime

data = [{"name": "aa", "created_at": "2022-01-17 07:38:26.403Z", "status": "success"},
        {"name": "bb", "created_at": "2021-12-1 07:38:26.403Z", "status": "failed"},
        {"name": "kk", "created_at": "2022-01-13 07:38:26.403Z", "status": "success"},
        {"name": "ll", "created_at": "2021-12-17 07:38:26.403Z", "status": "success"},
        ]
df = pd.DataFrame(data)
df
    name    created_at  status
0   aa  2022-01-17 07:38:26.403Z    success
1   bb  2021-12-1 07:38:26.403Z     failed
2   kk  2022-01-13 07:38:26.403Z    success
3   ll  2021-12-17 07:38:26.403Z    success

Build new column

def date_convert(x):
    x = x.split('.')[0]
    return datetime.strptime(x, "%Y-%m-%d %H:%M:%S").strftime("%B")


df['month_name'] = df.created_at.apply(date_convert)
df
    name    created_at  status  month_name
0   aa  2022-01-17 07:38:26.403Z    success     January
1   bb  2021-12-1 07:38:26.403Z     failed  December
2   kk  2022-01-13 07:38:26.403Z    success     January
3   ll  2021-12-17 07:38:26.403Z    success     December

Calculate & Build New DataFrame

group_month = df[df.status == "success"].groupby('month_name').groups
data = [{"created_at": i, "success": len(j)} for i, j in group_month.items()]
new_df = pd.DataFrame(data)
new_df
    created_at  success
0   December    1
1   January     2