I have recently been studying probability generating functions, and have seen the proof that the sum of two independent Poisson random variables has a Poisson distribution also. This used the fact that $G_{X+Y}(t)=G_{X}(t)G_{Y}(t)$ for independent random variables $X$ and $Y$.
I then thought 'What would the generating function of $X-Y$ be?". Since $X-Y=X+(-Y)$, we can use the above result if we can find the p.g.f. of $-Y$ (or, more generally, $cY$ for some constant $c$).
Since $-Y$ takes values $0,-1,-2\ldots$ with respective probabilities $p_{0},p_{1} \ldots$, substituting this into the definition of a generating function (or at least the definition I have) gives $$G_{-Y}(t)=\sum_{k=0}^{\infty}p_{k}t^{-k}=G_{Y}\left(\frac{1}{t}\right)$$
And a similar argument shows $G_{cY}(t)=G_{Y}(t^{c})$. Is this correct? My textbook doesn't say anything about this, and I can't find it on the internet either.
Note that $G_{cX}(t)=E(t^{cX})=\sum_{k=0}^{\infty}t^{cx}\cdot p_k$.
Now let $t^c=m$.
Then we have $G_{cX}(t)=\sum_{k=0}^{\infty}m^x\cdot p_k=G_{X}(m)$.
Again $m=t^c$.
So $G_{cX}(t)=G_{X}(t^c)$
Clearly your argument is right and here it is proved.