Let $f(z) =e^\frac1z + \frac{1}{z-1}$.
I need to determine what kind of singularities (removable, essential or pole) occur at $z=0$ and $z=1$.
My understanding of them is that at $z=0$: $$f(z) =e^\frac1z - 1$$ Which would make this an essential singularity in the same way that it is for $f(z) =e^\frac1z$, correct?
And for the singularity at $z=1$, this leaves us with: $$f(z) =e+\frac{1}{z-1}$$ Which would make this singularity a pole, right?
You are right, but for the wrong reasons. You have $f(z)=e^{1/z}+\frac1{z-1}$ everywhere; in particular, it is not true that, near $0$, $f(z)=e^{1/z}-1$. However, near $0$ you have$$f(z)=\sum_{n=-\infty}^0\frac{z^n}{(-n)!}-\sum_{n=1}^\infty z^n,$$and therefore $f$ has an essential singularity at $0$.
Near $1$, $e^{1/z}$ has a Taylor series $\sum_{n=0}^\infty a_n(z-1)^n$, and therefore$$f(z)=(z-1)^{-1}+\sum_{n=0}^\infty a_n(z-1)^n.$$It follows from this that $f$ has a pole of order $1$ at $1$.