Simplify binary

Pyth, 20 16 bytes

u?-GTG`u+yNsTG0z

4 bytes thanks to Jakube

Half of the code (u+yNsTG0) is simply the base conversion code.

Test Suite

u?-GTG`u+yNsTG0z
                    z = input() (The string of 1s and 0s)
                    T = 10
u              z    Apply until the value stops changing, starting with z
                    G is the current value, a string of 0s and 1s.
 ?-GT               If G - T, e.g., G with the digits 1 and 0 removed is not empty,
     G              Return G, to end the iteration.
       u     G0     Else, reduce over G with initial value 0.
         yN         Double the running total
        +  sT       and add the next digit, cast to an int.
      `             Convert to string.

The input 1 is handled by the fact that u notices the value has stopped changing.

CJam, 24 23 bytes

q{:~{1$++}*s__,(As*-!}g

Try it online in the CJam interpreter.

How it works

q                        Read all input.
 {                   }g  Do:
  :~                       Evaluate each character. Maps '0' -> 0 and '1' -> 1.
    {    }*                Fold; for each integer but the first:
     1$                      Copy the second-topmost integer.
       ++                    Add all three integers on the stack.
           s__             Cast to string and push two copies.
              ,(           Calculate string length and subtract 1.
                As         Push the string "10".
                  *        Repeat the string length-1 times.
                   -       Remove its elements from the string representation
                           of the integer.
                    !      Apply logical NOT.
                         If `!' pushed 1, repeat the loop.

Pip, 28 27 bytes

Ta=1|aRMta:$+(^a)*2**RV,#aa

Takes input as a command-line argument. We want to loop until a=1 or a contains some character(s) besides 0's and 1's. This latter condition is tested by RM'ing all characters in t = 10 from a. If there's anything left, the condition is truthy.

Inside the loop, the conversion works as follows:

a:$+(^a)*2**RV,#a

              ,#a  range(len(a))
            RV     reversed
         2**       2 to the power of each element
    (^a)*          multiplied item-wise with each digit in split(a)
  $+               Sum
a:                 and assign back to a

Putting a at the end auto-prints it.

A recursive solution in 28 bytes:

a<2|aRMt?a(f$+(^a)*2**RV,#a)