I know that $x(2t) = \frac12x(t)$. What is the Laplace transformation of $x(2t)$? Can you show all steps?
$$ \begin{align} X(s) &= \int_{-\infty}^{+\infty} x(2t)e^{-st}\,\,dt \\ &= \frac12\int_{-\infty}^{+\infty} x(t)e^{-st}\,\,dt\\ &= ? \end{align} $$
Definition of Laplace transform is
$$ \mathcal{L}[x](s) = \int_{0}^{+\infty} x(t)e^{-st}dt = X(s). $$
If $a>0$ and $\left(S_{a}x\right)(t)=x(at)$ we have
$$ \mathcal{L}[S_ax](s) = \int_{0}^{+\infty} \left( S_a x \right)(t)e^{-st}dt = \int_{0}^{+\infty} x(at)e^{-st}dt = \frac{1}{a}\int_{0}^{+\infty} x(t)e^{-\frac{s}{a}t}dt = \frac{1}{a}\left(S_{\frac{1}{a}} X\right)(s) = \frac{1}{a} X\left(\frac{s}{a}\right), $$
which means the operator $S_a$ is mapped into $\frac{1}{a} S_\frac{1}{a}$. Therefore if $x(t)$ is transformed into $X(s)$ then $x(2t)$ will be transformed into $\frac{1}{2} X\left( \frac{s}{2}\right)$. But you also have $x(2t) = \frac{1}{2} x(t)$ hence you have the identity
$$ \frac{1}{2} X\left(\frac{s}{2}\right) = \frac{1}{2} X(s) \Rightarrow X\left(\frac{s}{2}\right) = X(s). $$
You can easily derive the recurrence
$$ X\left( \frac{s}{2^k} \right) = X(s) $$
and for $k \to +\infty$ you have
$$ X(0) = X(s) $$
which means the Laplace transform of your function is a constant, more specifically
$$ \mathcal{L}[x](s) = \int_{0}^{+\infty} x(t)dt $$
For the recurrence:
$$ X(s) = X\left(\frac{s}{2}\right) = X\left(\frac{s}{2^2}\right) = \ldots = X\left(\frac{s}{2^k}\right) $$
For the continuity:
Let $s_k$ be a sequence such that $|s_k| \to 0$, we have
$$ \lim_{k \to \infty} X(s_k) = \lim_{k \to \infty} \int_{0}^{+\infty} x(t)e^{-s_kt} dt = \int_{0}^{+\infty} x(t) \left(\lim_{k \to \infty} e^{-s_k t} \right) dt = \int_{0}^{+\infty} x(t) dt = X(0), $$
of course this works if $X(0)$ exists.