Change of argument in Laplace Transform 
The Laplace transform of a function $f=f(x)$ has the following definition:
  $$\mathcal{L}({f(x)})=\int^{\infty}_0e^{-sx}f(x)dx \tag{1}$$

However, when $f=f(ax-c)$, where $a$ and $c$ are arbitrary constants, we have
$$\mathcal{L}(f(ax-c))=\int^{\infty}_0e^{-sx}f(ax-c)dx \tag{2}$$
with which I got very confused, why should $(2)$ not be
$$\mathcal{L}(f(ax-c))=\int^{\infty}_0e^{-s(ax-c)}f(ax-c)d(ax-c) \tag{3}$$
or equivalently $$\mathcal{L}(f(ax-c))=a\int^{\infty}_0e^{-s(ax-c)}f(ax-c)dx \tag{3}$$
instead?

The Laplace transform of a function $f(x)$ is itself not a function of $x$. 
You can write $$\mathcal{L}(f(x))(s) = F(s) = \int_{0}^\infty e^{-s\tau} f(\tau) d\tau$$
where $\tau$ is a dummy index of integration. 
So when you write $g(x) = f(ax - c)$, the Laplace transform $G(s)$ of $g(x)$ is \begin{align*}
G(s) &= \int_{0}^\infty e^{-s \lambda} g(\lambda) d\lambda\\
&= \int_{0}^\infty e^{-s \lambda} f(a \lambda - c) d\lambda
\end{align*}
