How about a noncommutative extension of R? If σ is a ring automorphism on R, let xc = σ(c)x, and build the formal power series as before. This is called a twisted ring. The maximal ideal is generated by M and x, and anything outside this ideal, starting with a unit in R, is a unit in S. Combine σ with synthetic division to find the inverse. I'll leave the details to you.
Let f be the function that turns every group element into e and collapses the resulting terms. Verify that f is a ring homomorphism from S onto K. It's kernel is the linear combinations of group elements where the coefficients sum to 0. Call this ideal J. It's going to become the jacobson radical - hence the letter J.
We will prove that every y outside of J is a unit. That will do the trick.
Let's start with something easier - let G be a p cycle. The automorphisms of G move the generator c to any other nontrivial group element, e.g. c → c7. This induces a ring automorphism of KG fixing K. Does it fix anything else? The coefficient on c has to agree with the coefficient on c7, and c14, and c21, and so on. All the coefficients, outside of the constant term, must agree. Divide through by this coefficient and get q+c+c2+c3+…, for some q in K.
Raise this to the p power. Since all the terms commute, the frobenius homomorphism applies, and we can raise each term to the p power separately. The result is qp + (p-1), or qp-1. Use the frobenius homomorphism again and rewrite this as (q-1)p. This is nonzero, as long as q is not 1. Therefore, our original expression, fixed by the automorphisms of KG, is a unit iff its coefficients sum to something nonzero in K.
Given y outside of J, multiply y by all its conjugates, giving z. The coefficients of Z sum to something nonzero in K, and z is fixed by the automorphisms of KG, hence z is a unit, and y is a unit. Therefore J is the jacobson radical of KG.
Now let G be a p group. The center of G is a nontrivial p group, and contains a p cycle, generated by some c in G. Let H be the factor group G/{c}. Thus KH is a quotient ring of KG. Let y in KG lie outside of J. In other words, the coefficients of y add up to something nonzero. The same holds for the image of y in KH. By induction on the size of the p group, KH is local, and the image of y is a unit in KH. Find z such that yz maps to 1 in KH. In other words, yz is a linear combination of powers of c, such that the coefficients sum to 1. We just prove that such an expression is a unit in the group ring of c over K, which makes it a unit in KG. Therefore y is a unit in KG, and KG is local, with maximal ideal J.
Such a ring need not be reduced either. Let K = G = Z2. If c generates G, (1+c)2 = 0. This can be generalized to characteristic p.
When G is abelian, and K is a field, all of J is nil. Take an expression in J and raise it to the p, by raising each term to the p. Do this again and again until the group elements become trivial. The coefficients are then added together, and that is the same as adding them together first, and raising the sum to a power of p. The result is 0. J is the nil radical, and there are no prime ideals other than J.