Suppose $X_i$ (where $X_i$ are i.i.d exponential random variables with parameter $\lambda$) is the time it takes for someone to complete a task. Each attempt is successful with prob $p$. We assume independence for each attempt and independence of the durations.
I have modeled the total time as the following:
$S_N = X_1 +X_2 +...X_n$ where $P(N=n) = p(1-p)^{n-1}$. Additionally the moment generating function for each $X_i$ is given by $M_{X_i}(t) = \frac{1}{1-\lambda{t}}$.
Is is the incorrect intuition to just proceed from the following:
$M_{S_N}(t)$ = ($\prod{M_{X_i}})*p(1-p)^{n-1}$
We observe that $$M_S(t) = \operatorname{E}[e^{tS}] = \operatorname{E}[\operatorname{E}[e^{tS} \mid N]],$$ where the outer expectation is taken with respect to $N$ and the inner with respect to $S$ conditioned on $N$. Since $$e^{tS} \mid N = \prod_{i=1}^N e^{t X_i},$$ we have $$\operatorname{E}[e^{tS} \mid N] \overset{\text{ind}}{=} \prod_{i=1}^N \operatorname{E}[e^{tX_i}] = \left(M_X(t)\right)^N = \left(\frac{\lambda}{\lambda-t}\right)^N,$$ where $X$ is an exponential random variable parametrized by rate $\lambda$. Consequently, $$M_S(t) = \operatorname{E}\left[e^{N \log (\lambda/(\lambda - t))}\right] = M_N\left(\log \frac{\lambda}{\lambda - t}\right),$$ where $N \sim \operatorname{Geometric}(p)$ with parametrization $$\Pr[N = n] = (1-p)^{n-1} p, \quad n = 1, 2, \ldots.$$ We require $N$ to have support beginning at $1$ since it is not possible to correctly answer a question in $N = 0$ responses. This parametrization has MGF $$M_N(t) = \frac{pe^t}{1 - (1-p)e^t},$$ from which we obtain $$M_S(t) = \frac{\frac{p\lambda}{\lambda - t}}{1 - \frac{(1-p)\lambda}{\lambda - t}} = \frac{p\lambda}{p\lambda - t},$$ which is the MGF of an exponential distribution with rate $p\lambda$, as desired.