@TobiasDiez: I see what you mean now. The proof that I had in mind looks at the geometry of the $k$-jet space $J^k(M,\mathbb{R}^3)$, where $M$ is a surface, with its natural action of the $6$-dimensional isometry group $G$ of Euclidean $3$-space. What we are asking for is a $G$-invariant function $\Phi:J^k(M,\mathbb{R}^3)\to\mathbb{R}^3$ such that two immersions $f_1,f_2:M\to\mathbb{R}^3$ are congruent under $G$ if and only if $\Phi(j^k(f_1))=\Phi(j^k(f_2))$. The proof that such a $\Phi$ does not exist involves some knowledge of exterior differential systems.

@TobiasDiez: I'm not sure what you mean by 'one cannot use differential invariants to characterize an immersion'. Are you referring to the fact that one can have two non-congruent, smooth unit speed space curves with torsion $\tau\equiv0$ and equal curvatures $\kappa$? I gave a simple example of such in an answer to an earlier question (mathoverflow.net/q/428800).

@IgorKhavkine: However, that answer has something wrong with it. The statement 'If a vector bundle admits a flat connection, then the rational Pontryagin classes of the tangent bundle vanish..." is just false. I suspect that I.B. meant to start with "If a tangent bundle admits a flat connection...", which would have been a true statement. Also, the final statement that 'most vector bundles don't admit a flat connection' is true, but that says nothing about the vector bundles over particular manifolds, such as $G/\Gamma$.

@IgorKhavkine: The OP has not asked that $E$ be the tangent bundle of $G/\Gamma$. Indeed, the tangent bundle of $G/\Gamma$ is trivial, so it does have a flat connection.

You may be missing some hypotheses. Otherwise, there are very simple examples provided by smooth complex line bundles with nontrivial first Chern class over $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$.

When you write "the moduli space of connections", do you mean "the space of connections modulo gauge equivalence"?

There certainly is something wrong with the second claim: If $S\equiv0$ and $\mathrm{dim}(M)>1$, then $M_S = M$ is open and connected (if $M$ is connected), but the eigenvalues of $S$ are not distinct. I think you are being careless about the statement of the second property. I think it should be something like "The $E_S$ distinct eigenvalues of $S$ on a connected component of $M_S$ are smooth".