Z-Transform of a^|n|
I believe the right way to decompose this function is:

$f(n) = a^n u(n)  +  a^{-n} u(-n)$

Then the answer for your question is:

$Z( f(n) ) = \frac{Z}{Z-a} + \frac{Z^{-1}}{Z^{-1} - a}$
EDIT: Looks like it is wrong answer. :(



Ok, let's start from the beggining and try to find right answer step by step

Given sequence is $f(n) = a^{|n|}$ . 'a' is a positive constant.$

1) Let's  decompose it into sum of 2 sequences:

$f(n) = a^n u(n)  +  a^{-n} u(-n-1)$

2) If we define $f_1(n) = a^n u(n)$  and $f_2(n) = a^{-n} u(-n-1)$
Then $Z( f(n) ) = Z( f_1(n) )+Z( f_2(n) )$



3) $Z( f_1(n) ) = \sum_{n=-\infty}^{\infty} a^n u(n) z^{-n} = \sum_{n=0}^{\infty} (az^{-1})^n = \frac{1}{1-az^{-1}} = \frac{z}{z-a}  $



4) $Z( f_2(n) ) = \sum_{n=-\infty}^{\infty}  a^{-n} u(-n-1) z^{-n} = \sum_{n=-\infty}^{-1} (az)^{-n} = \sum_{n=1}^{\infty} (az)^{n} = -1+1+\sum_{n=1}^{\infty} (az)^{n} = -1+\sum_{n=0}^{\infty} (az)^{n} = -1 + \frac{1}{1-az} = \frac{-1+az+1}{1-az} = \frac{az}{1-az} $



5) $Z( f(n) ) = Z( f_1(n) )+Z( f_2(n) ) = \frac{z}{z-a} + \frac{az}{1-az} = \frac{z}{z-a} - \frac{z}{z-\frac{1}{a}} $

If there is anything unclear or wrong in this derivation, please comment.

As for wolfram answer - I don't know why it is so. I never used wolfram before. 
But it perfoms symbolic computations and you must be sure that your assumptions about range of n (does $n\in[-\infty,\infty]$  or $ n\in[0,\infty]$ ) for example should coinside with wolfram assumptions.