take a single element of the series
$f(t)=\delta(t-k T)$
I used a sampling time $T$ in the expression. You can set this so 1 if you like.
The Laplace transform of $\delta(t)$ is $1$
The Laplace transform of $g(t-kT)$ is $e^{-skT}G(s)$
so
$\large \mathscr{L} \{ \delta(t-kT) \} = e^{-s k T} * 1 = e^{-s k T}$
The Laplace transform is linear so the Laplace transform of the sum is the sum of the transforms, i.e.
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} f_k(t) \} = \displaystyle{\sum_{k=0}^\infty} F_k(s)$ where $F_k(s) = \mathscr{L}\{ f_k(t) \}$
thus
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} \delta(t - kT) \} = \displaystyle{\sum_{k=0}^\infty} e^{-s k T}$
This last term is a geometric series in $k$
$\large \displaystyle{\sum_{k=0}^\infty} e^{-s k T} = \dfrac 1 {1 - e^{-s T}}$
so
$\large \mathscr{L} \{ \displaystyle{\sum_{k=0}^\infty} \delta(t - kT) \} = \dfrac 1 {1 - e^{-s T}}$