The Laplace transform of $\frac{d}{dt} f(t)$ would be sF(s), when f(0)=0, which is something you can find in a Laplace transform table.
If there is a rule that prohibits mathematical operations from $v_{\text{L}}(t) = L \frac{di_{\text{L}}(t)}{dt}$ to $\frac{ V_{\text{L}} (t) } {i_L (t)} = L \frac{d}{dt}$ and do the Laplace transform after that to make the result of this Laplace transform equal to sL, because $i_L(t)$ is not a constant, what would be wrong with this reasoning: The laplace transform of $v_L(t)$ is $V_L(s)$, the Laplace transform of $i_L(t)$ is $I_L(s)$, the Laplace transform of $V_{\text{L}}(t) = L \frac{di_{\text{L}}(t)}{dt}$ is $V_{\text{L}}(s) = L \cdot s \cdot I_L(s)$, so $\frac{V_L (s)}{I_L (s)} = L \cdot s $ and the inverse Laplace transform of $\frac{V_L (s)}{I_L (s)} = L \cdot s $ is $\frac{ V_{\text{L}}(t) } {i_L (t)} = L \frac{d}{dt}$?
No: Recalling the quotient rule $$\frac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{\frac{df}{dx}g(x)-f(x)\frac{dg}{dx}}{g(x)^2}=\dfrac{1}{g(x)}\dfrac{df}{dx}-\frac{f(x)g'(x)}{g(x)^2},$$ we conclude that $\dfrac{1}{g(x)}\dfrac{d}{dx}f(x) \neq \dfrac{d}{dx}\dfrac{f(x)}{g(x)}$ unless $g(x)=$ const.