This is a comment, not an answer, but is too long to fit in the comment box: having read the question, answers, and comments, I don't quite follow the intent of this question:

We can define a manifold to be a locally ringed space in which each point has a neighbourhood
isomorphic to an open subset of ${\mathbb R}^n$ (or even just ${\mathbb R}^n$ itself) with its sheaf of smooth functions (plus second countability and  Hausdorffness, if you like).    As was remarked by Dmitri, the collection of all such will then form an atlas, but one doesn't need to say this.

As Pete Clark says, what I've said so far is evident.

But it seems that another aspect of the question is whether one can always avoid working in coordinates.  This seems to have nothing to do with atlases.

E.g. in arguments in the locally ringed space set-up, one will certainly in many instances verify that a property can be checked locally, and then verify it on Euclidean space with
its natural smooth structure.  (Just as in the theory of schemes, one often shows that a property is local, and then checks it in the affine case.)   

Now one can ask: can one avoid the latter kinds of arguments?  This seems unlikely: manifolds are <I>defined</I> to be locally Euclidean (no matter which of the possible formalisms one is using), and so if one is proving theorems about manifolds, one will
have to use this somewhere.  For example, one can surely define the tangent sheaf in
a coordiante free way, but to prove that it is locally free of rank equal to the dimension
of the manifold, one is going to reduce to a local computation and then appeal to calculus
on Euclidean spaces; there is no other way!

[EDIT: The last sentence may be too categorical of a declaration; see Dmitri Pavlov's answer for a suggestion of a more substantially algebraic reformulation of the
notion of manifold.]