Does this shuffle have non-zero probability for all permutations?

Question

I was trying to do some code golf, when I created the following algorithm to shuffle a string:

for(;k-->0;y=s,s="")for(var c:y.split(""))s=Math.random()<.5?s+c:c+s;

To explain better what it does, I've recreated it in Python:

import random
str = "Randomize me!"
print('[',str,']',sep='')
temp=""
for _ in range(len(str)): # repeat as many times as the string is long
  for char in str:        # take the string's chars and randomly construct a temporary
    temp = temp + char if random.choice([True, False]) else char + temp
  temp, str = "", temp    # use the temporary as the basis for the next iteration
print('[',str,']',sep='')

Also, per request, here it is in pseudocode:

SHUFFLE(STR)
 0 TEMP ← ""
 1 LOOP STR.LENGTH TIMES:
 2     FOR EACH CHAR ∈ STR:
 3         IF COIN-FLIP
 4             TEMP = TEMP + CHAR
 5         ELSE
 6             TEMP = CHAR + TEMP
 7         END IF
 8     END FOR
 9     STR ← TEMP
10     TEMP ← ""
11 END LOOP
12 RETURN STR

The output looks randomized to me. However, if you only consider one iteration of the algorithm, the first character put into temp is always str[0]. The second character put into temp on this iteration is always str[1], which must be next to str[0] according to the way characters are appended. Therefore, after one iteration, the probability that the shuffled string has the first character outside of the center is 0, and the probability that it is not adjacent to the second character is 0.
I tried to remedy this problem by repeating the shuffle once for each character in the string. Intuitively, it seems to work, but I can't seem to prove that all permutations of the input string can show up.

So, here's the question: do all permutations of the input string have non-zero probability as output from this shuffle?