The inverse of function $\text{erf}(\sqrt{s}\cdot x)$ is relatively easy to obtain:
$$\mathcal{L}_s^{-1}\left[\text{erf}\left(\sqrt{s}\cdot x\right)\right](t)=\delta (t)-\frac{|x| \cdot \theta \left(t-x^2\right)}{\pi t \sqrt{t-x^2}}$$
But the inverse of $\text{erfi}(\sqrt{s}\cdot x)$ seems to be much more complicated.
Let the integrand from Mellin's inverse formula be:
$$F(s,x,t)=\text{erfi}\left(\sqrt{s} x\right) e^{s t}$$
Hint: The integration path along the imaginary axis from $(-i\infty, i\infty)$ results in very complicated expressions including the Meijer G-function.
Hence we integrate around the branch cut in 3 steps:
Integrate below the real axis from $(-\infty,0]$: $$I1=\frac{1}{2\pi i}\int_{-\infty }^0 F(\xi -i k,x,t) \, d\xi=-\frac{i \left(e^{-i k t} \text{erfi}\left(\sqrt{-i k} x\right)-\frac{x \left(\text{erfi}\left(\sqrt{-i k \left(t+x^2\right)}\right)+i\right)}{\sqrt{t+x^2}}\right)}{2 \pi t}, t>0$$
Integrate along the imaginary axis $[-k i, k i]$: $$I2=\frac{1}{2\pi}\int_{-k}^k F(i \omega ,x,t) \, d\omega=\frac{i \left(\frac{x \left(\text{erfi}\left(\sqrt{i k \left(t+x^2\right)}\right)-\text{erfi}\left(\sqrt{-i k \left(t+x^2\right)}\right)\right)}{\sqrt{t+x^2}}+e^{-i k t} \text{erfi}\left(\sqrt{-i k} x\right)-e^{i k t} \text{erfi}\left(\sqrt{i k} x\right)\right)}{2 \pi t} $$
Integrate above the real axis from $[0,-\infty)$: $$I3=\frac{1}{2 \pi i}\int_0^{-\infty } F(\xi +i k,x,t) \, d\xi=-\frac{i \left(\frac{x \left(\text{erfi}\left(\sqrt{i k \left(t+x^2\right)}\right)-i\right)}{\sqrt{t+x^2}}-e^{i k t} \text{erfi}\left(\sqrt{i k} x\right)\right)}{2 \pi t}$$
The final sum evaluates to: $$I1+I2+I3=\fbox{$-\frac{x}{\pi t \sqrt{t+x^2}}\text{ if }t>0$}$$
Unfortunately we have a result for $t>0$, but not for $t=0$.
Let Mathematica tell us the laplace transform of our result:
$$\mathcal{L}_t\left[\fbox{$-\frac{x}{\pi t \sqrt{t+x^2}}\text{ if }t>0$}\right](s)=\text{erfi}\left(\sqrt{s} x\right)-\frac{2 \log (x)}{\pi }$$
So here we have an additional term $\frac{2 \log (x)}{\pi }$ which corresponds to $\frac{2 \delta (t) \log (x)}{\pi }$ in time domain.
What is the correct solution for $\mathcal{L}_s^{-1}\left[\text{erfi}\left(\sqrt{s} x\right)\right](t)$ including $t=0$?
My suggestion: $$\mathcal{L}_s^{-1}\left[\text{erfi}\left(\sqrt{s}\cdot x\right)\right](t)=\frac{2\log (x)}{\pi }\delta (t)-\frac{|x|}{\pi t \sqrt{t+x^2}} $$
Is that result correct?
Any help is highly appreciated, thanks.