An alternative way is to use the propagator. I will set $\hbar^2/m=1$ so that the propagator is: $$ \begin{align} G &= \int e^{-itk^2/2}e^{ikx}\frac{dk}{2\pi} \\ &= \frac{e^{ix^2/2t}}{\sqrt{-i2\pi t}} \end{align} $$ Setting $a=1$, your wave function is: $$ \begin{align} \psi &= \int e^{-|y|}\frac{e^{i(x-y)^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= \int_{-\infty}^x e^{y-x}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy+\int_x^{+\infty} e^{x-y}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= -\frac{e^{-x+it/2}}{\sqrt{-i2\pi t}}\int_{-x}^{+\infty} e^{i(y+it)^2/2t}dy+\frac{e^{x+it/2}}{\sqrt{-i2\pi t}}\int_x^{+\infty} e^{i(y+it)^2/2t}dy \\ &= -\frac{e^{-x+it/2}}{2}\text{erfc}\left(\frac{it-x}{\sqrt{-i2\pi t}}\right)+\frac{e^{x+it/2}}{2}\text{erfc}\left(\frac{it+x}{\sqrt{-i2\pi t}}\right) \end{align} $$

You can recover this formula by residue calculus, but you will need to recognise the inverse Fourier transform of the error function.

An alternative way is to use the propagator. I will set $\hbar^2/m=1$ so that the propagator is: $$ \begin{align} G &= \int e^{-itk^2/2}e^{ikx}\frac{dk}{2\pi} \\ &= \frac{e^{ix^2/2t}}{\sqrt{-i2\pi t}} \end{align} $$ Setting $a=1$, your wave function is: $$ \begin{align} \psi &= \int e^{-|y|}\frac{e^{i(x-y)^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= \int_{-\infty}^x e^{y-x}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy+\int_x^{+\infty} e^{x-y}\frac{e^{iy^2/2t}}{\sqrt{-i2\pi t}}dy \\ &= -\frac{e^{-x+it/2}}{\sqrt{-i2\pi t}}\int_{-x}^{+\infty} e^{i(y+it)^2/2t}dy+\frac{e^{x+it/2}}{\sqrt{-i2\pi t}}\int_x^{+\infty} e^{i(y+it)^2/2t}dy \\ &= -\frac{e^{-x+it/2}}{2}\text{erfc}\left(\frac{it-x}{\sqrt{-i2\pi t}}\right)+\frac{e^{x+it/2}}{2}\text{erfc}\left(\frac{it+x}{\sqrt{-i2\pi t}}\right) \end{align} $$

You can recover this formula in Fourier space. However, as Hyperon remarked, your approach was wrong because you are neglecting the contribution of the arc that closes the contour. In your case Jordan's lemma does not apply, and in fact it is precisely this arc that will give you the error function.