If σ has a nontrivial kernel, then for c in that kernel, xc = 0. In other words, x is a zero divisor. This is usually not what we want, so for the rest of this page, assume σ is injective.
If R is a domain, consider the lowest term in the product of two polynomials or two series. This looks like c*σ(d), which is always onnzero. Therefore the twisted extension of a domain is a domain.
Now consider a polynomial p with nilpotent coefficients. If p has l terms, expand pn to find ln products, as each of l terms is placed in each of n positions. The trick is to choose n high enough. Let m be the highest exponent on any of the l coefficients of p. Then set n = mlk. At least mk instances of one of the terms of p will appear in each product. Pull all these coefficients to the start, and at least m will be run through the same power of σ. When σj(c) is raised to the m, the result is 0, just like cm. Each product drops to 0, and p is nilpotent.
Adjoin any number of indeterminants, and select a σ for each one, as it commutes with R, such that each σ has a finite period. The above proof remains valid, though the exponent must be much higher, to accommodate all the different ways a coefficient might run through the various automorphisms that are introduced by the (finitely many) indeterminants in a given polynomial. I'll leave the details to you. With these stipulations in mind, nilpotent coefficients produce nilpotent polynomials.
The converse is true in commutative rings, but not here. Let R = Z[u,v], with uv = 0. Let σ swap u and v. Now ux is a nilpotent polynomial, even though u is not nilpotent in R.
Ad vx, another nilpotent polynomial, and get (u+v)x. Rais this to the nth and get (un+vn)xn. Thus the sum of nilpotent polynomials need not be nilpotent.
If p is a nilpotent polynomial, of any form, apply σ to all the coefficients and find another nilpotent polynomial with the same exponent. This is because σ respects all the ring operations that take place when p is multiplied by itself again and again.
In general, σ extends to an endomorphism, and in this case a monomorphism, on the entire twisted ring. If σ is surjective on R then it is surjective on the twisted ring. Thus an automorphism extends to an automorphism.
If a polynomial or series is a unit, its constant coefficient must be a unit.
If R is a domain, the lead terms in two polynomials combine to create a nonzero term in the product. Therefore the units of the extension are precisely the units of R.
Like any other ring, 1-w is a unit when w is nilpotent. Divide 1 by 1-w using synthetic division, and the process stops.
In the commutative world, a unit polynomial is always some unit in R plus a nilpotent polynomial - the converse of the above. This is not the case here. We don't even need a twist to demonstrate this fact; simply choose an appropriate noncommutative base ring R. Start with Z and adjoin the indeterminant u, giving the polynomials in u. Then adjoin 1/u, so that u is a unit in R. Adjoin the indeterminant v, such that u and v do not commute. Then let vn = 0. Now 1-vx is a unit, since vx is nilpotent. Multiply by u to get another unit, u-uvx. But uvx is not nilpotent, since u and v do not commute.
Let σ be an R endomorphism, building the twisted series over x. (Here σ need not be injective.) As mentioned earlier, a series has to start with a unit to be a unit. Premultiply by the inverse of this unit, so that our series f(x) starts with 1. Divide this into 1 by synthetic division. The quotient, g(x), begins with 1. Subtract 1*f(x) from 1, and the linear term of the "remainder" is -a1x. Multiply this by f(x) on the right and subtract. The remainder begins with a squared term, and the process continues forever. This builds g(x), such that g*f = 1. By symmetry, f is also right invertible, hence f is a unit. A series is a unit iff its constant term is a unit.