Show that two $R$-algebras are not isomorphic. Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.
Does anyone have any pointers on how to approach this?
I don't know a purely algebraic proof. The following proof involves a little algebraic geometry and algebraic topology, so it may not be the intended answer to your question. In that case, please specify your context (i.e. where did you get the problem from).
Suppose $A$ and $B$ are isomorphic as $\Bbb R$-algebras. Then their corresponding $\Bbb R$-points, viewed as real manifolds, are isomorphic, as polynomial maps are continuous (and even $C^\infty$).
However, these two manifolds are: $A(\Bbb R) = \Bbb R^4 \backslash \{0\}$ and $B(\Bbb R) = \operatorname{GL}(2, \Bbb R)$. It is obvious that they are not isomorphic as topological spaces, by e.g. comparing their simplicial (co)homology groups.
Question: &quot;Does anyone have any pointers on how to approach this?&quot;
Answer: Your argument is correct: Define
$$A := \mathbb{R}[a,b,c,d]/(a^2+b^2+c^2+d^2), B := \mathbb{R}[a,b,c,d]/(ad-bc).$$
Formally, if there is an isomorphism of affine schemes (over $k$ - the real numbers) $\phi: T:=Spec(A) \cong S:=Spec(B)$ it would follow that $T$ and $S$ have the same $k$-rational points: The map $\phi$ would induce a 1-1 correspondence
$$\phi(k): T(k) \cong S(k)$$
and an equality of sets
$$T(k) \cong Hom_{k-alg}(A,k) \cong Hom_{k-alg}(B,k)\cong S(k),$$
but $T$ has only one $k$-rational point: $x:=(0,0,0,0)$. $S$ has a rational point $x:=(a,b,c,d)$ for any $x$ with $det(x)=0$.
Formally: If $T\cong S$ are isomorphic schemes (over $k$) it follows $h_T\cong h_S$
are isomorphic as functors (over $k$). Hence an isomorphism $\phi$ of affine schemes over $k$ does not exist.