It is enough to do the case in which $r=1.$
Let $X$ be the number of independent trials before the first success, with probability $p$ of success on each trial. Then $X\in\{0,1,2,3,\ldots\}$ and $\Pr(X=x) = (1-p)^x p.$ Then we have $$
\begin{align}
M_X(t) & = \operatorname E(e^{tX}) = \sum_{x=0}^\infty e^{tx} (1-p)^x p \\[8pt]
& = \frac p {1 - (1-p)e^t} = \frac{1-q}{1-qe^t}.
\end{align}
$$
Now let $X$ be the number of independent trials needed to get one success, with probability $p$ of success on each trial. Then $X\in\{1,2,3,\ldots\}$ and $\Pr(X=x) = (1-p)^{x-1} p.$ Then we have $$
\begin{align}
M_X(t) & = \operatorname E(e^{tX}) = \sum_{x=1}^\infty e^{tx} (1-p)^{x-1} p \\[8pt]
& = \frac{pe^t}{1 - (1-p)e^t} = \frac{(1-q)e^t}{1 - qe^t}.
\end{align}
$$
Thus the one on the MGF page is correct if you consider the "negative binomial distribution" supported on the set $\{r,r+1,r+2,\ldots\},$ and the one on the NB page is correct if you consider the NB distribution supported on the set $\{0,1,2,3,\ldots\},$ and the roles of $p$ and $q$ are interchanged.
Notice that Wikipedia's article titled "Negative binomial distribution" says this:
Different texts (and even different parts of this article) adopt slightly different definitions for the negative binomial distribution. They can be distinguished by whether the support starts at $k = 0$ or at $k = r,$ whether $p$ denotes the probability of a success or of a failure, and whether $r$ represents success or failure, so it is crucial to identify the specific parametrization used in any given text.