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Python 3, 30 26 24 bytes

Extremely basic:

lambda s:s.count('\x65')

4 bytes off thanks to @Challenger's suggestion (adapted to this solution).

51 46 44 bytes, but I like it better:

lambda s:''.join('1'for i in s if i=='\x65')

Output is in unary. I found it first in the hopes that the other way was impossible; for once I'm annoyed that Python lets me do things the easy way.

5 bytes off thanks to @Challenger.

2 bytes off of each thanks to @xnor.

Attribution for the simple method: https://stackoverflow.com/questions/1155617/count-occurrence-of-a-character-in-a-string

Attribution for the unary one: https://stackoverflow.com/questions/1450897/python-removing-characters-except-digits-from-string