I want to create a shell that extracts the difference that existed the other day and does not exist today by using the diff command.

There is a command to classify two sorted files, the comm command, into lines that exist only in each file and lines that are common.(If you're not particular about using the diff command...)
The comm command must give a file with the lines sorted.(From the text of the question, it seems like a file that can be sorted, but you need to check if it is.)

First, use the following command:
for parts that exist only in A.txt, parts that exist only in B.txt, and parts that are common. I can separate them, so please check them out.

comm<(sort A.txt)<(sort B.txt)

You can specify options to print only the three parts that you want.A. Specify --23 を to output only parts that exist in txt.

comm-23<(sort A.txt)<(sort B.txt)

Difference
that existed the other day (B) and does not exist today (A) 「I didn't know what the condition of "ignore the presence or absence of A" meant.

List of recent days B.txt

d
c
B
a

Today's List A.txt

a
c
e

In , b and d no longer exist, and e appears new.
Do you mean you want to extract b and d at this time?

If so,
diff<(sort A.txt)<(sort B.txt)

1a2
>b
3c4
<e
---
>d

of
You can extract lines that begin with >.

[Procedure]

diff<(sort A.txt)<(sort B.txt)|grep'^>'|sed's/^>//'

[Results]

b
d

[Environment of execution]
GNU bash, version 4.4.20(1)-release(x86_64-pc-linux-gnu)
diff(GNU diffutils) 3.6
grep(GNU grep) 3.1
sed(GNUsed) 4.4
sort (GNU coreutils) 8.28