I am trying to find the Laurent series of the function $ f(z) =\frac{1}{\exp (z) -1 }$ above but I don't know how to find the term
$$a_{-1} = \frac{1}{2\pi i}\int f(\xi ) \,d\xi $$
I tried parameterising but I ended up with a tricky integral so I was wondering if there was an easier way to evaluate the above integral
You really do have to specify where you want to expand the Laurent Series.
For example, if you want to find the poles and corresponding residues of $f$, you might want to expand the series centered at each isolated singularity.
The singularities of $f$ are where $e^z-1=0$, i.e. $z=2k\pi i$ for $k \in \Bbb Z$.
The Taylor Series for $e^z$ centered at $2k\pi i$ is given by
$$e^z=1+(z-2k\pi i)+\frac 12 (z-2k\pi i)^2 + \cdots = \sum_{n=0}^ \infty \frac{1}{n!}(z-2k\pi i)^n$$
Thus the Laurent Series is
\begin{align} f(z) & =\frac{1}{e^z-1} \\ & = \biggl ((z-2k\pi i)+\frac 12 (z-2k\pi i)^2 + \cdots \biggr)^{-1} \\ & = (z-2k\pi i)^{-1}-\frac 12 (z-2k\pi i)^0+\cdots \end{align}
Hence the $\text{res}(f;2k\pi i)=1$.
As for your integral, you have not specified the set (path) on which you are integrating over.
But since we have found the poles and residues, it follows easily by Cauchy's Residue Theorem that
$$\int_\gamma f(z)dz=2 \pi n i$$
for any closed, continuously differentiable path $\gamma$ that does not contain any point $z=2k \pi i$, and where $n$ is the number of poles inside $\gamma$.