If the S-matrix has symmetry group $G$, must the fields be representations of $G$? If the fields in QFT are representations of the Poincare group (or generally speaking the symmetry group of interest), then I think it's a straight forward consequence that the matrix elements and therefore the observables, are also invariant.
What about the converse:

If I want the matrix elements of my field theory to be invariant scalars, how do I show that this implies that my fields must be corresponding representations? 

How does this relate to S-matrix theory?
 A: Fields aren't "representations of the Poincare group".  Fields -- some fields, anyways, like scalar and Dirac spinor fields (gauge fields are more complicated)  -- take values in vector spaces which are representations of the Lorentz group.  This does not guarantee that observables are 'invariant'; observables can and do change their values when you act by a symmetry.  Spin up can change to spin down if you rotate your coordinate system on its head.
This sort of confusion is what comes of the sloppy notational habits of the theoretical physics community.  Learn you some representation theory, for great justice.
A: The question formulation (v2) seems to mix the notions of invariant and covariant, which essentially is also the main point of user1504's answer (v1).
Let's say we have a group $G$. The group $G$ could e.g. be a finite group or a Lie group. When we say that a theory is invariant under $G$, it normally implies at least two things. 


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*The group $G$ acts on the theory. This in particular means that there is a well-defined given prescription on how the constituents of the theory change under the action of the group. Often in physics (but not always), it happens that the group action is linearly realized, i.e. the fields, the matrix elements, and other objects form linear representations of the group $G$. At this stage, the representations could be reducible or irreducible, finite dimensional or infinite dimensional. The corresponding object then behaves covariantly (not necessarily invariantly) under the action of the group. If a representation is completely reducible, we can decompose it in irreps.

*
*

*In case of an off-shell formulation: The action $S$ is off-shell invariant under the group $G$. Or phrased equivalently, the action $S$ form a trivial representation of the group.

*In case of an on-shell formulation: The equations of motion behave covariantly under the group $G$.



A refinement. Ron Maimon makes in a comment an important point that if there is a hierarchy of say two theories, and if the group action is a priori only defined in the smaller theory, then the group $G$ does not necessarily have to have a well-defined action on the larger theory. For instance,


*

*
*

*Small theory = on-shell formulation; 

*Large theory = off-shell formulation.


*
*

*Small theory = minimal $S$-matrix formulation; 

*Large theory = an underlying field theoretic formulation.


*
*

*Small theory = formulation on gauge-invariant physical subspace/submanifold/phase space;

*Large theory = formulation on BRST-extended space/manifold/phase space.


