They are not isomorphic as $k$-algebras, at least. (From a geometric perspective, this is the more natural question....)
The first ring -- say $R$ -- is isomorphic to $k[t]$, so is the ring of regular functions on the affine line $\mathbb{A}^1$ over $k$. In other words, it corresponds to the projective line minus a single point at infinity.
The second ring -- say $S$ -- is the ring of polynomial functions on the unit circle. Geometrically, it is the projective line minus two points at infinity. If $\sqrt{-1} \in k$, these points are both rational and $S \cong k[t,t^{-1}]$. Thus the unit group of $S$ is larger than just $k^{\times}$ and so they cannot be isomorphic as $k$-algebras.
In case $\sqrt{-1}$ does not lie in $k$, if they were isomorphic as $k$-algebras then by tensoring with $k(\sqrt{-1})$ they would also be isomorphic as $k(\sqrt{-1})$-algebras and one can reduce to the previous case. But actually in this case one can do better: the Picard group of $S$ has order $2$, whereas the Picard group of $R$ is trivial, so $R$ and $S$ are not isomorphic even as rings. (See e.g. page 6 of this paper for a treatment of the Picard group of $S$.)
This answer leaves open the possibility that $R \cong S$ as rings when $\sqrt{-1} \in k$. Probably someone else will address this...
Added: The third ring $T = k[x,y]/(y^2-x^2(x+1))$ is the coordinate ring of an affine curve which is singular at the origin, so is not even a Dedekind domain, unlike $R$ and $S$.