We can define it as \newcommand\Laplace[1]{\ensuremath{\mathscr{L}\mleft\{#1\mright\}}} where the \mleft\{ and \mright\} will make the braces scale appropriately with taller arguments.
\documentclass{article}
\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{mleftright}
\newcommand\Laplace[1]{\ensuremath{\mathscr{L}\mleft\{#1\mright\}}}
\begin{document}
\begin{align*}
\Laplace{\delta(t)}&=1\\
\Laplace{\frac{1}{(n-1)!}t^{n-1}e^{-at}}&=\frac{1}{(s+a)^n}\\
\end{align*}
\end{document}

It might be useful to define an inverse laplace command as well: \newcommand\InvLaplace[1]{\ensuremath{\mathscr{L}^{-1}\mleft\{#1\mright\}}}
EDIT: Thanks to @Mico for recommending \mleft and \mright from mleftright for removing the extra space between the Laplace operator and the start of the braces.
\{\}after you type\Laplace(why?) or are you asking for a tex defintion, and if so what do you want it to do? – David Carlisle Nov 16 '21 at 21:29\newcommand\Laplace[1]{\mathscr{L}\{#1\}}then\Laplace{A}would make ℒ{A} ? – David Carlisle Nov 16 '21 at 21:31