I hope I don't make a complete fool of myself, but here is my try:

From the example I understood that it is OK to trash INDF.
I interpreted 'including return' as including the 2 cycles for the return.

decf  INDF, w
bsf   INDF, 7
andwf INDF, w
return

the formula is ( x - 1 ) & ( x | 0x80 )

explanation:
(x-1) leaves all set bits in place, except the rightmost one,
which is 'smeared out' over the lower bits.
Hence (x-1)&(x) is 0 when there was only one bit set, OR x was 0.
From x==0 follows (x-1) == 0xFF, and (x-1) & (anything<>0) != 0.
When only one bit is set the highest bit in (x-1) will be clear,
so it has no effect (except in the all clear case, wher
it has the desired effect) to use (x|0x80).

verification:

#include <stdio.h>

int f( int x ){
   return ( 0xFF & ( (x - 1) & ( x | 0x80 ) ) );
}

int ones( int x ){
   int n = 0;
   while( x > 0 ){
      if( x & 1 > 0 ){
         n++;
      }
      x = x >> 1;
   }
   return n;
}

int main( void ){
   int i, j, k;
   for( i = 0; i < 256; i ++ ){
      j = f( i );
      k = ones( i );
      printf( "%2X %2X %2X %d\n", i, j, k, ( k == 1 ) == ( j == 0 ) );
   }
   return 0;
}

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