I know it's been a while since this question was asked, but there is (at least now there is) a one-liner that is supported by the documentation:
In [4]: df
Out[4]:
label
0 (a, c, e)
1 (a, d)
2 (b,)
3 (d, e)
In [5]: df['label'].str.join(sep='*').str.get_dummies(sep='*')
Out[5]:
a b c d e
0 1 0 1 0 1
1 1 0 0 1 0
2 0 1 0 0 0
3 0 0 0 1 1