I don't immediately see a way to use the first identity. I will attempt to show by definition of Laplace transform.

[math]
\displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \int_0^\infty e^{-st} \left(1 - \dfrac2{\sqrt \pi} \int_0^{\frac a{2\sqrt t}} e^{-x^2}\,dx\right)\,dt = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \int_0^{\frac a{2\sqrt t}} e^{-st} e^{-x^2}\,dx\,dt
[/math]

We change the order of integration of the double integral:

[math]
\displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \int_0^{\frac{a^2}{4x^2}} e^{-st} e^{-x^2}\,dt\,dx = \dfrac1s - \dfrac2{\sqrt \pi} \int_0^\infty \dfrac{1 - \exp\left(-\dfrac{a^2s}{4x^2}\right)}s e^{-x^2}\,dx
[/math]

Expand and we have

[math]
\displaystyle\mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} =\dfrac1s - \dfrac2{s\sqrt \pi} \int_0^\infty e^{-x^2}\,dx + \dfrac2{s \sqrt \pi} \int_0^\infty \exp\left(-x^2 - \dfrac{a^2s}{4x^2}\right)\,dx = \dfrac2{s \sqrt \pi} \int_0^\infty \exp\left(-x^2 - \dfrac{a^2s}{4x^2}\right)\,dx = \dfrac2{s\sqrt \pi} \int_0^\infty \exp\left(-\left(x - \dfrac{|a|\sqrt s}{2x}\right)^2 - |a| \sqrt s\right)\,dx
[/math]

So

[math]
\displaystyle \mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac{2e^{-|a|\sqrt s}}{s\sqrt \pi} \int_0^\infty \exp\left(-\left(x - \dfrac{|a|\sqrt s}{2x}\right)^2\right)\,dx
[/math]

We are left with an integral that can be solved easily with the right trick. Use the

Cauchy-Schlömilch transformation here and we get

[math]
\displaystyle \mathcal L\left\{\operatorname{erfc}\left(\dfrac{a}{2 \sqrt t}\right)\right\} = \dfrac{2e^{-|a|\sqrt s}}{s\sqrt \pi} \int_0^\infty \exp\left(-x^2\right)\,dx
[/math]

The rest is trivial.