Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.
Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)
In fact, the simplest example of this kind is given by primary Kodaira surfaces (http://en.wikipedia.org/wiki/Kodaira_surface), they have two holomorphic $1$-forms.
These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf
To get the Dolbeault complex, you need a choice of holomorphic structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty_M\otimes_{\mathcal{O}_M}\mathcal{E}$ and
$\overline{\partial}_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$
is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying
$$ \overline{\partial}_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}_E(\sigma),$$
for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that
$\left. \overline{\partial}_E\right| _{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$.
You can then extend the Dolbeault operator to
$\overline{\partial}_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$,
$\overline{\partial}_E^2=0$,
by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution
$$ 0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$
The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.
You cannot do any of this without the holomorphic structure. Differently put, you need
the total space of $E$ to be a complex manifold and the projection $E\to M$ to be holomorphic.
You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.
ADDENDUM
The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure
$\overline{\partial}_E$ is non-trivial. It can be phrased as follows:
$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that
$D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.
Best Answer
A. Moroianu gives a detailed proof on pp. 72-74 of his Lectures on Kähler geometry (Theorem 9.2), available on the internet. (The preprint has it as Theorem 3.2.)
He attributes that proof to S. Kobayashi, Differential geometry of complex vector bundles. I guess he means Proposition I.3.7 there, which Kobayashi couches in a different language involving connections.
P.S. Note that Moroianu gives himself a whole complex of operators $\bar\partial_E:\Omega^{p,q}(E)\to\Omega^{p,q+1}(E)$, and Donaldson-Kronheimer at least the $(0,q)$ ones, plus Leibniz. I'm not 100% clear that just the $(0,0)$ one plus Leibniz suffice to determine everything, as the question seems to imply.