At least two ways of looking at this:
The Laplace representation of the capacitor's reactance is \$\frac{1}{sC}\$, hence for a voltage, \$\small V(s)\$ across \$\small C\$, the current through \$\small C\$, by Ohm's law, will be \$\small I(s)=sC\:V(s)\$
Differentiation in the time domain is equivalent to multiplying by \$\small s\$ in the Laplace domain. Therefore \$\frac{dv}{dt}\small \rightarrow sV(s)\$, and the differential equation transforms to: \$\small I(s) = C\:sV(s)\$
Zero initial conditions has been assumed. For example, a fully charged open-circuit capacitor will have zero current, but non-zero voltage.