One may define the derivative of $f$ at $x$ as $\lim\limits_{h\to0}\cdots\cdots\cdots$ etc., and show that that has certain properties, but it also has a "physical" interpretation: it is an instantaneous rate of change.
How much money do I need to put in the bank today to have $\$1$, $t$ years from now, assuming continuous compounding at a constant rate? The answer is $(e^{-st} \times \$1)$ where $s$ is the annual interest rate. So how much do I need to deposit today to get paid at a rate of $f(t)$ in dollars per year, $t$ years from now? The answer is $$ \int_0^\infty e^{-st} f(t)\,dt. $$ This is a "physical" interpretation of the Laplace transform as the "present value" of a revenue stream, as a function of the interest rate.
Is that pretty much the whole story of how to "physically" interpret the Laplace transform, or can more be said?