Question: I am doing a self-study using Artin's algebra. There is a problem which asks us to prove that adjoining an element to a ring doesn't do anything if that element was already in the ring.

More formally, say that $\alpha=a$ where $a\in R$. We want to show that $R\approx R[\alpha]$. (Where $\approx$ means "isomorphic to".) 

My proof: We can consider $R[\alpha]\approx R[x]/(x-\alpha)$, which intuitively means "killing" $x-a$ in the ring of polynomials. Any function $f\in R[x]$ can be written as $(x-\alpha)g(x)+r(x)$ for some g,r by the division algorithm. Since the degree of r must be less than the degree of $x-a$, r must be a constant polynomial. Since $(x-\alpha)=0$, the residue of $f$ in $R[x]/(x-\alpha)$ is just r, which is some element of R. We can see that every $s\in R$ has a corresponding constant polynomial $f(x)=s$ in $R[x]$, which means the simple inclusion map $s\mapsto f(x)=s$ is our desired isomorphism.

So, this proof seems fine to me, except that I seem to have proven that adjoining *any* element keeps the ring the same. This sounds implausible to me. Where did I go wrong?

Answer: 
Your proof doesn't show that adjoining any element keeps the ring the same, because if $\alpha$ isn't in $R$, then $x-\alpha$ is not in $R[x]$, so the quotient $R[x]/(x-\alpha)$ doesn't make sense.
How have you defined $R[\alpha]$? I usually define it as the smallest ring containing $R$ and $\alpha$, in which case, if $\alpha$ is in $R$, it's clear that $R[\alpha]=R$ (and if $\alpha$ is not in $R$, it's clear that $R[\alpha]\ne R$).