There are certain functions that can be non-differntiable but are still solution to a "differential equations".
This is fine, as a non-differentiable function can still have an integral.
The main question is the following:
What does it mean by taking the Laplace transform of Non-differntiable functions.?
Isnt it a violation of some rules here and there?
Laplace transform is an integral and works for any integrable function does not matter if its differentiable or not with a certain Region of Convergence region. Your question to be put precise i think is what happens when you do, $L^{-1} (s \times L(f(x)))$ where $f(x)$ is non-differentiable. The thing is laplace transform being an integral based transform allows impulses $\delta(x)$ as functions which will allow differentiation of a lot more functions. For the functions like $|x|$ where the derivative just does not converge, Laplace transform also does not give proper results at those points.
So $L^{-1} (s \times L(f(x)))$ will differentiate $f(x)$ properly but only except for the points of $x$ where left derivative and right derivative does not converge. It will put values at these non-differentiable points to make sure that transform and inverse transform is consistent i.e., the integral equation of transform works.
Further since measure of these discrete set of points is $0$. So when you use Laplace transform, the solution is unique only upto a countably many points of $x$. So we impose continuity constraint on the solution and make it unique.