Isomorphism of the affine circumference over certain fields. Let us consider the coordinate ring of the circumference
$$
A:=K[X,Y]/(X^{2}+Y^{2}-1),
$$ 
and let us suppose that $K$ is infinite but not necesarilly algebraically closed. I wonder if $A$ is isomorphic to 
$$
K[X]\text{ or } S^{-1}K[X],
$$
where $S=\{1,X,X^{2},\ldots\}$. I am not sure, but the answer could be different if $-1$ is a square in $K$ or not.
 A: For convenience sake, let's assume that $\mathrm{char}(K)\ne 2$.
To analyze your situation, consider the smooth projective model
$$C:X^2+Y^2=Z^2\subseteq\mathbb{P}^2_K$$
of your $\text{Spec}(A)$. Note that $C$ is a smooth projective curve of genus $1$. It's rational (i.e. isomorphic to $\mathbb{P}^1_K$) if and only if it has a point. 
If $K=\mathbb{R}$ then $C$ does have a $\mathbb{R}$-point (thanks to the OP for pointing out a silly mistake!). Thus, we know that $C\cong\mathbb{P}^1_\mathbb{R}$. Note though that the complement of $\text{Spec}(A)$ in $C$ is $V(X^2+Y^2)$ which is a degree $2$-point on $C$ (corresponding to the $\text{Gal}(\mathbb{C}/\mathbb{R})$-orbit $[\pm i:1:0]$ in $C_\mathbb{C}$). Thus, regardless, as we mention below, $\text{Spec}(A)=C-\{V(X^2+Y^2)\}$ is not isomorphic to $\mathbb{A}^1_\mathbb{R}$.
But, let's suppose that $i\in K$ (i.e. that $K$ has a root of $-1$). Then, note that $C$ DOES have a point: $[1:i:0]$. Thus, $C\cong\mathbb{P}^1_K$. Moreover, note that $C-\text{Spec}(A)$ is just 
$$V(X^2+Y^2)=[\pm i:1:0]$$
So, we see that $\text{Spec}(A)$ is $\mathbb{P}^1_K$ minus two $K$-rational points, so it's $\mathbf{G}_m=\text{Spec}(K[t,t^{-1}])$. Thus, in this case we see that it's not isomorphic to $K[t]$.
This last result is to be expected in some sense. Colloquially $A$ looks like the ring of functions on a circle (e.g. $\text{Spec}(A)$ is a $K$-group scheme, and the Lie group one gets as $\text{Spec}(A)(\mathbb{R})$ when $K=\mathbb{R}$ is, in fact, $S^1$).
Anyways, the above this actually shows that $K[t]$ is NEVER isomorphic to $A$ since, if they were isomorphic, they'd be isomorphic over $\overline{K}$ which, as we mentioned in the previous paragraph, they're not.
I'll leave it to you as an exercise to think what happens in characteristic $2$ (if you really want :) )
