I certainly didn't interpret your original question as asking for a proof. Consider the definition of a binomial coefficient when one of the values is negative:

$$\binom{-n}{m} = (-1)^m \binom{m+n-1}{m}$$

See Wiki - Binomial coefficient.

So when $n=1$ we have $\binom{-1}{m}=(-1)^m$. That should be enough to show the two equations are equal. (And there is the appearance that you really weren't looking for a proof but rather if there was an inconsistency in WolframAlpha. No problem in that. There are some inconsistencies. But it would be better to state that explicitly. You'll get more and better help that way.)

@Jim Baldwin I can obtain the same result with the simple built-in function Binomial[n,m]. Maybe is not possible to "prove" that Binomial[-1,m]=-1 for each m natural number.

In other words, I'd like to "prove" that the following two solutions are equal (see result):

http://www.wolframalpha.com/input/?i=InverseZTransform%5B1%2F(1-a*(z%5E-1))%5D+and+InverseZTransform%5Bz%2F(z-a)%5D

I suppose you know that all of the following statements are true:

Binomial[-1, 0] == 1
Binomial[-1, 1] == -1
Binomial[-1, 2] == 1
Binomial[-1, 3] == -1
Binomial[-1, 4] == 1

So while I don't know why you would want to change the usual definition, you could always use a replacement rule at the end of your calculations:

x + 2 Binomial[-1, m] /. Binomial[-1, m_] -> -1
(* -2+x *)

Hi @Jim Baldwin , m is a generic natural number.