The Wikipedia article states the asymptotics of $E_{a,b}(z)$ for $z\to-\infty$, but the argument given there, for when and why it holds, isn't clear for me (regarding why, it seems to rely on the behaviour of $E_{a,b-ka}(z)$ as $z\to-\infty$, which looks like a circular dependence). A rigorous treatment may be based on the contour integral representation, which is proved using the Hankel's $\frac{1}{\Gamma(s)}=\frac{1}{2\pi\mathrm{i}}\int_\lambda z^{-s}e^z\,dz$ (where the contour $\lambda$ encircles the negative real $z$-axis, and the principal value of $z^{-s}$ is taken), valid for all complex values of $s$, and Watson's lemma (or its complex extension if needed).
Specifically, for real $x>0$ and $0<a<1$, we have (again with the p.v. of $z^a$) $$f_a(x):=x^{a-1}E_{a,a}(-x^a)=\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{e^{xz}\,dz}{1+z^a};\tag{*}$$ making $\lambda$ encircle closely the negative real axis, in the limit of "closely", we get $$f_a(x)=\frac{1}{2\pi\mathrm{i}}\int_0^\infty\left(\frac{1}{1+e^{-\mathrm{i}\pi a}t^a}-\frac{1}{1+e^{\mathrm{i}\pi a}t^a}\right)e^{-xt}\,dt$$ (substituting $z=e^{\pm\mathrm{i}\pi}t$ on the upper/lower edge of $\lambda$ with $+/-$ respectively). For $t<1$, we have $$\frac{1}{2\pi\mathrm{i}}\left(\frac{1}{1+e^{-\mathrm{i}\pi a}t^a}-\frac{1}{1+e^{\mathrm{i}\pi a}t^a}\right)=\frac1\pi\sum_{n=1}^{\infty}(-1)^{n-1}t^{na}\sin n\pi a,$$ and a variant of Watson's lemma (for a power series in $t^a$) gives $$f_a(x)\asymp\frac1\pi\sum_{n=1}^{(\infty)}(-1)^{n-1}\sin n\pi a\,\frac{\Gamma(na+1)}{x^{na+1}}=\bbox[5px,border:2px solid]{\sum_{n=1}^{(\infty)}\frac{(-1)^n}{\Gamma(-na)}\frac{1}{x^{na+1}}.}$$
This gives what you've been told for $0<a<1$. At $a=1$, the entire series vanish, and in fact many things change, starting with a note that $(\textrm{*})$ still holds if $\lambda$ encircles the disc $|z|\leqslant 1$, not only the negative real axis (in this case, $a>1$ is also admissible); the asymptotic analysis similar to the above is still possible, but it must be done more carefully. The same pertains to complex values of $x$ (however, still with $\Re x>0$).
Not sure what you mean by "principle part". I think you mean the Laurent series, which is going to be different for each singularity of a function.
I think you are mixing up a Laurent series vs a rational series. A Laurent series is of the form:
$$f(z) = \sum_{k=-\infty}^{\infty} a_k(z-z_0)^k $$
Notice how the series is centered at $z_0$, which is a singularity to the meromorphic function. A rational series is
$$ f(z) = \sum_{k=0}^\infty \frac{a_k}{z - b_k}$$
Notice that there is no "center". This is not a Laurent series, does not have an annulus of convergence and is instead governed by different theorems. However, you can use any sequence of convergent analytic functions to discover properties about the Laurent series.
For example, if I wanted coefficient $a_{-1}$ in the Laurent series centered at $n$, for the function defined by the rational series
$$f(z) = \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n}$$
Then all I have to do is integrate around a small compact contour $C$ that is centered at $n$, canceling out those terms that integrate to zero:
$$\begin{align} \int_C \sum_{k=-\infty}^{\infty} a_k (z-n)^k &= \int_C \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n} \\ a_{-1}\cdot 2\pi i &= \int_C \frac{1}{z-n} \\ a_{-1}\cdot 2\pi i &= 2\pi i \\ a_{-1} &= 1 \end{align}$$
So now we know that the Laurent series centered at $n$ has a principle coefficient of $1$. Again, note there is a different Laurent series for each singularity.
Best Answer
Well $\;\displaystyle zE_{a,b+a}(z)=\sum_{h=0}^{\infty}\frac{z^{h+1}}{\Gamma((h+1)a+b)}$ so yes it is immediate.
For other interesting properties of the Mittag-Leffler function see this thread.