There actually IS a way to do it in Python 2.3+, but it's a bit esoteric. I don't know if you realize this, but you can do the following:
$ unzip -l /tmp/example.zip
Archive: /tmp/example.zip
Length Date Time Name
-------- ---- ---- ----
8467 11-26-02 22:30 jwzthreading.py
-------- -------
8467 1 file
$ ./python
Python 2.3 (#1, Aug 1 2003, 19:54:32)
>>> import sys
>>> sys.path.insert(0, '/tmp/example.zip') # Add .zip file to front of path
>>> import jwzthreading
>>> jwzthreading.__file__
'/tmp/example.zip/jwzthreading.py'
According to the zipimport library:
Any files may be present in the ZIP archive, but only files .py and .py[co] are available for import. ZIP import of dynamic modules (.pyd, .so) is disallowed. Note that if an archive only contains .py files, Python will not attempt to modify the archive by adding the corresponding .pyc or .pyo file, meaning that if a ZIP archive doesn't contain .pyc files, importing may be rather slow.
Thus, all you have to do is zip the files up, add the zipfile to your sys.path and then import them.
If you're building this for UNIX, you might also consider packaging your script using this recipe: unix zip executable, but note that you might have to tweak this if you plan on using stdin or reading anything from sys.args (it CAN be done without too much trouble).
In my experience performance doesn't suffer too much because of this, but you should think twice before importing any very large modules this way.