One way to remember it is that if you think of the integral in the inverse formula as a sort of sum (completely non-rigorous here):
$ f(t) = c_1 e^{s_1 t} + c_2 e^{s_2 t} + c_3 e^{s_3 t} + \cdots $
Here the coefficient $c_1$ is proportional to the Laplace transform at $s_1$, and so on. Then since the derivative of $e^{st}$ with respect to $t$ is $se^{st}$, you get that the coefficients in the sum are multiplied by $s$ when you take the derivative:
f'(t) = c_1 s_1 e^{s_1 t} + c_2 s_2 e^{s_2 t} + c_3 s_3 e^{s_3 t} + \cdots
Conversely, integrating $f$ multiplies the coefficients by $1/s$. Again, this is not in any way a proof, but hopefully it gives some intuition.
Another way to think about it is that the smoothness of a function is related to its decay in the frequency domain as $s \to \infty$. Differentiating makes a function less smooth, (a twice differentiable function becomes onces differentiable, etc.) while integrating makes a function more smooth (a bounded discontinuous function becomes continuous, etc.). So it makes sense that differentiation would make a function decay slower in the frequency domain, in the sense that $1/s$ decays slower than $1/s^2$.