The Laplace transform of the complementary error function $\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)$ is
$$L\left\{\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)\right\}=L\left\{1-\operatorname{erf}\left(\frac{1}{\sqrt{t}}\right) \right\}$$
That is
$$\begin{aligned} L\left\{\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)\right\} &=L\left\{1-\frac{2}{\sqrt{\pi}} \int_{0}^{1/\sqrt{t}} e^{-x^2} dx \right\}\\ &=L\left\{1-\frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1/\sqrt{t}} x^{2n} dx \right\} \\ &=L\left\{1-\frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} \frac{1}{t^{n+1/2}} \right\} \\ &=\frac{1}{p} - \frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} L\left\{\frac{1}{t^{n+1/2}} \right\} \\ \end{aligned}$$
After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!
It looks fine. Now you just have to exploit $$\mathcal{L}(1)=\frac{1}{s},\qquad \mathcal{L}\left(\frac{1}{t^{n+1/2}}\right) = s^{n-\frac{1}{2}}\Gamma\left(\frac{1}{2}-n\right)=s^{n-\frac{1}{2}}(-1)^n\frac{2^n\sqrt{\pi}}{(2n-1)!!}. $$ Anyway, It would have been faster to exploit the properties of the Laplace transform. $\text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that $$ \mathcal{L}\left(\text{Erfc}\left(\frac{1}{\sqrt{t}}\right)\right) = \color{red}{\frac{e^{-2\sqrt{s}}}{s}}. $$