Solving the Definite Integral $\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t$

Represent the erf as an integral and work a substitution. To wit, the integral is

$$\frac{2}{\sqrt{\pi}} \int_0^1 dv \, \int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} $$

To evaluate the inner integral, we sub $y=a/t+v^2 t$. Then the reader can show that

$$\int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} = 2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y}$$

The latter integral is easily evaluated using the sub $y=2 v \sqrt{a} \cosh{w} $ and is equal to

$$2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y} = 2 \int_0^{\infty} dw \, e^{-2 v \sqrt{a} \cosh{w}} = 2 K_0 \left ( 2 v \sqrt{a} \right )$$

where $K_0$ is the modified Bessel function of the second kind of zeroth order. Now we integrate this expression with respect to $v$ and multiply by the factors outside the integral to get the final result:

$$\begin{align} \int_0^{\infty} dt \, t^{-3/2} e^{-a/t} \operatorname{erf}{\left ( \sqrt{t} \right )} &= \frac{4}{\sqrt{\pi}} \int_0^1 dv \, K_0 \left ( 2 v \sqrt{a} \right ) \\ &= 2 \sqrt{\pi} \left [K_0 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{-1}\left ( 2 \sqrt{a} \right ) + K_1 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{0}\left ( 2 \sqrt{a} \right ) \right ] \end{align}$$

where $\mathbf{L}$ is a Struve function.

You can also follow your Laplace approach. Define $$ I(\alpha)=\int_0^{\infty}\frac{\sin(2\alpha\sqrt{s})}{s\sqrt{1+s}}ds $$

now set $s=q^2$

$$ I(\alpha)=2\int_0^{\infty}\frac{\sin(2\alpha q)}{q\sqrt{1+q^2}}ds $$

Now differentiate with respect to $\alpha$

$$ I'(\alpha)=4\int_0^{\infty}\frac{\cos(2\alpha q)}{\sqrt{1+q^2}}ds $$

this integral now furnishs a representation of the modified Besselfunction $K_0(z)$

$$ I'(\alpha)=4 K_0(2\alpha ) $$

according to 10.43.2 backintegrating w.r.t. to $\alpha$ yields

$$ I(\alpha)=2\pi \alpha (K_0(2\alpha )L_{-1}(2\alpha )-K_1(2\alpha )L_{0}(2\alpha ))+C $$

where $L_{\nu}(z)$ are modified Struve function. The constant of integration $C$ is fixed to be zero by the condition $I(0)=0$. Multiplying with $1/\sqrt{\pi}\alpha$ yields the result obtained by @Ron Gordon