How to see that the shift $x \mapsto (x-c)$ is an automorphism of $R[x]$?

Hmmm...what about directly? $$f(x)=\sum_{n=0}^\infty a_nx^n\,,\,g(x)=\sum_{n=0}^\infty b_nx^n\,\in R[x]$$ with $\,\,a_n=b_n=0 \,\,\text{ for all but a finite number of indices}$ . Since $$f(x)+g(x)=\sum_{n=0}^\infty (a_n+b_n)x^n\,\,,\,f(x)g(x)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)x^n$$ we get that, denoting the map by$\,\phi(x):=x-c\,$, we have $$\phi(f+g)=\sum_{n=0}^\infty (a_n+b_n)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n+\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)+\phi(g)$$ $$\phi(fg)=\sum_{n=0}^\infty\left( \sum_{k=0}^n a_kb_{n-k}\right)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)\phi(g)$$

Remember: all the above are in fact finite sums, so we can re-arrange them as we wish. Also, $\,\phi(r)=r\,\,\,\forall\,r\in R\,$, so in particular $\,\phi(1)=1\,$ .

As Lubin noted in a comment above, the first thing to check is that the map which sends $p(x)$ to $p(x-c)$ is indeed a ring homomorphism. To do this, we have the following tool:

Theorem: Let $R$ be a commutative ring. The polynomial ring $R[X]$ satisfies the following universal property: given a commutative ring $A$ containing $R$ and an element $a$ of $A$, there is a unique ring homomorphism $$\phi:R[X] \to A$$ such that $\phi(X)=a$.

(Note that there exists a similar version for the polynomial ring in finitely many variables.) Hence, by this universal property there exists a unique ring homomorphism that sends $X$ to $X-c$, for any $c\in R$. And as azarel noted above, this is in fact a ring isomorphism, since you have an inverse (namely, $X\mapsto X+c$).

Hint: The map $x\to x+c$ is his inverse.