I am looking for the inverse Laplace transform of $$\frac{1}{s-1}e^{-\sqrt{s}x}$$ This is for an introductory partial differential equations class so I am thinking that it should not be too hard. If it wasn't for the $s-1$ in the denominator and we had just $s$ instead, then the transform would simply be $$\textrm{erfc} \left(\frac{x}{2\sqrt{t}}\right) $$
Edit: Here is the problem from where that transformation originated. 
You could simply convolve the ILT with that of $1/(s-1)$, which is $e^t$. But I prefer using the definition of the ILT in the complex plane.
In this case, consider the contour integral
$$\oint_C dz \frac{e^{-x \sqrt{z}}}{z-1} e^{t z}$$
where $C$ is a Bromwich contour in the left half-plane with a detour along the negative real axis and around the branch point at $z=0$, as pictured below.
(The pole at $z=1$ is represented by the small disk.)
The contour integral is equal to, taking the limits as the outer radius goes to $\infty$
$$\int_{c-i \infty}^{c+i \infty} \frac{ds}{s+1} e^{-x \sqrt{s}} e^{s t} + \int_{\infty}^0 \frac{dy}{y+1} e^{-i x \sqrt{y}} e^{-t y} + \int_0^{\infty} \frac{dy}{y+1} e^{i x \sqrt{y}} e^{-t y}$$
The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=1$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s+1} e^{-x \sqrt{s}} e^{s t} = e^{-x} e^t - \frac1{\pi} \int_0^{\infty} dy \, \frac{\sin{x \sqrt{y}}}{1+y} e^{-t y} $$
To evaluate the integral, sub $y=u^2$ and get
$$\int_0^{\infty} dy \, \frac{\sin{x \sqrt{y}}}{1+y} e^{-t y} = \int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t u^2} - \int_{-\infty}^{\infty} \frac{du}{1+u^2} \frac{\sin{x u}}{u} e^{-t u^2}$$
The first integral may be evaluated using Parseval's theorem:
$$\int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t u^2} = \frac1{2 \pi} \int_{-x}^{x} dk \, \pi \, \sqrt{\frac{\pi}{t}} e^{-k^2/(4 t)} = \pi \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )}$$
The second integral may be expressed as an integral of the first. Write that integral as
$$\begin{align}e^t \int_{-\infty}^{\infty} \frac{du}{1+u^2} \frac{\sin{x u}}{u} e^{-t (1+u^2)} &= e^t \int_t^{\infty} dt' \, e^{-t'} \, \int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t' u^2} \\ &= \pi\, e^t \int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} \end{align}$$
Now I will state the result here for now so the forest isn't lost for the trees:(*)
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} + \frac12 e^x \operatorname{erfc}{\left (\sqrt{t}+\frac{x}{2 \sqrt{t}} \right )} - \frac12 e^{-x} \operatorname{erfc}{\left (\sqrt{t}-\frac{x}{2 \sqrt{t}} \right )} $$
and thus
ADDENDUM
I will now derive the result (*). Begin by integrating by parts:
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} - \frac{x}{2 \sqrt{\pi}} \int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')}$$
$$\underbrace{\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')}}_{t'=1/u^2} = 2 \int_0^{1/\sqrt{t}} du \, e^{-1/u^2} e^{-x u^2/2} = 2 e^{-x} \int_0^{1/\sqrt{t}} du \, e^{-\left (\frac1{u} - \frac{x}{2} u \right )^2}$$
Now sub $v=\frac1{u} - \frac{x}{2} u$. Then
$$u = \frac1{x} \left (\sqrt{v^2+2 x}-v \right ) $$ $$du = \frac1{x} \left (\frac{v}{\sqrt{v^2+2 x}}-1 \right ) dv $$
and carrying out the substitution...
$$\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')} = \frac{2 e^{-x}}{x} \int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, \left (\frac{v}{\sqrt{v^2+2 x}}-1 \right ) e^{-v^2}$$
The first term evaluates to
$$\begin{align}\int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, \frac{v}{\sqrt{v^2+2 x}} e^{-v^2} &= \frac12 \int_{\infty}^{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}}\right )^2} dw\, (w+2 x)^{-1/2} e^{-w}\\ &= \frac12 e^{2 x} \int_{\infty}^{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}}\right )^2} dw\, w^{-1/2} e^{-w} \\ &= e^{2 x} \int_{\infty}^{\sqrt{t} + \frac{x}{2 \sqrt{t}}} dy \, e^{-y^2} \\ &= -\frac{\sqrt{\pi}}{2} e^{2 x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \end{align}$$
The second term is obviously simpler:
$$\int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, e^{-v^2} = -\frac{\sqrt{\pi}}{2} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} $$
Therefore...
$$\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')} =\frac{\sqrt{\pi}}{x} \left [e^{-x} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} - e^{x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \right ]$$
and...
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} - \frac12 \left [e^{-x} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} - e^{x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \right ]$$
ADDENDUM II
A little manipulation puts the answer in a slightly nicer form: