import difflib

>>> a = 'alex is a buff dude'
>>> b = 'a;exx is a buff dud'
>>> difflib.SequenceMatcher(None, a, b).ratio()

0.89473684210526316

If you don't have to use difflib, you could use a set and string splitting!

>>> original = "Apple Microsoft Google Oracle"
>>> edited = "Apple Nvdia IBM"
>>> set(original.split()).symmetric_difference(set(edited.split()))
{'IBM', 'Google', 'Oracle', 'Microsoft', 'Nvdia'}

You can also get the shared members with the .intersection()

>>> set(original.split()).intersection(set(edited.split()))
{'Apple'}

The Wikipedia has a good section on basic set operations with accompanying Venn diagrams
https://en.wikipedia.org/wiki/Set_(mathematics)#Basic_operations

However, if you have to use difflib (some strange environment or assignment) you can also just find every member with a +- prefix and slice off the all the prefixes

>>> diff = d.compare(original.split(), edited.split())
>>> list(a[2:] for a in diff if a.startswith(("+", "-")))
['Nvdia', 'IBM', 'Microsoft', 'Google', 'Oracle']

All of these operations result in an iterable of strings, so you can .join() 'em together or similar to get a single result as you do in your Question

>>> print("\n".join(result))
IBM
Google
Oracle
Microsoft
Nvdia

In [1]: from itertools import takewhile
In [2]: ''.join(a for (a, b) in takewhile(lambda (a, b): a == b, zip(t1, t2)))
Out[2]: '20180628-1'

Here we zip two strings together and iterate on them while their prefixes are equal.

Upd. Equivalent solution not using itertools:

result = []
for a, b in zip(t1, t2):
    if a != b:
        break
    result.append(a)
return ''.join(result)

If you remove the []'s from your strings, and call .split() on them in the .compare() perhaps you'll get what you want.

import difflib

# define original text
# taken from: https://en.wikipedia.org/wiki/Internet_Information_Services
original = "IIS 8.5 has several improvements related"

# define modified text
edited = "It has several improvements related"

# initiate the Differ object
d = difflib.Differ()

# calculate the difference between the two texts
diff = d.compare(original.split(), edited.split())

# output the result
print ('\n'.join(diff))

Output

+ It
- IIS
- 8.5
  has
  several
  improvements
  related

For starters, you need to pass strings to difflib.SequenceMatcher, not files:

# Like so
difflib.SequenceMatcher(None, str1, str2)

# Or just read the files in
difflib.SequenceMatcher(None, file1.read(), file2.read())

That'll fix your error.

To get the first non-matching string, see the difflib documentation.