You can mask the indices matching NaN positions and ffill to keep the previous existing index:

df.index = df.index.to_series().mask(df['PersonEntityID'].isna()).ffill()

Output:

   PersonEntityID
0        324141.0
0             NaN
2        425544.0
2             NaN
4        434456.0

If you want a new value per unique ID:

df.index = pd.factorize(df['PersonEntityID'].ffill())[0]

If you want a unique value only when a new ID is found you can combine np.maximum.accumulate and factorize:

df.index = np.maximum.accumulate(pd.factorize(df['PersonEntityID'])[0])

Output:

   PersonEntityID
0        324141.0
0             NaN
1        425544.0
1             NaN
2        434456.0

Here is a summary of the differences with a modified input:

   PersonEntityID  consecutive  per_unique  per_new_unique
0        324141.0            0           0               0
1             NaN            0           0               0
2        425544.0            2           1               1
3             NaN            2           1               1
4        324141.0            4           0               1

Use factorize with forward filling missing values:

df.index = pd.factorize(df['PersonEntityID'].ffill())[0]

Or for general solution use:

s = df['PersonEntityID'].ffill()
df.index = pd.factorize(s.ne(s.shift()).cumsum())[0]

Output with duplicated values:

df['unique'] = pd.factorize(df['PersonEntityID'].ffill())[0]

s = df['PersonEntityID'].ffill()
df['dupl'] = pd.factorize(s.ne(s.shift()).cumsum())[0]
print (df)
   PersonEntityID  unique  dupl
0        324141.0       0     0
1             NaN       0     0
2        425544.0       1     1
3             NaN       1     1
4        324141.0       0     2