I have been struggling with this problem for a while. So far, this is what I have managed to obtain. I have little faith in my solution. If my answer is wrong, could someone give me a hint?
We know that the Laurent series of $e^{1/z}$ centred at $z=1$ will be of the form
$$e^{1/z}=\sum_{-\infty}^\infty a_n(z-1)^n = \left(...+\frac{a_{-2}}{(z-1)^2} +\frac{a_{-1}}{(z-1)^1} + a_{0} + a_1(z-1)+a_2(z-1)^2 + ... \right)$$
At $z=1$, we see that the function $e^{1/z}$ is defined and has no singularity. Thus $a_n = 0$ if $n < 0$. So it does not have any non-zero terms of negative exponent.
As for the radius of convergence, we note that the $z=0$ gives us an essential singularity, so the radius of convergence is $R = 1$.