Which commutative rings $A$ have the property that there is exactly one ring homomorphism $\mathbb{R} \to A$?
For the sake of a better name, let's call these rings "really unique" (pun intended). Clearly, $0$ is really unique. It is well-known that $\mathbb{R}$ is really unique. The class of really unique rings is closed under limits in $\mathbf{CRing}$. So for example, $\mathbb{R}^I$ for every index set $I$ is really unique.
The polynomial ring $\mathbb{R}[X]$ is really unique: If $f : \mathbb{R} \to \mathbb{R}[X]$ is a ring homomorphism, then $f(r)$ is a unit for every $r \in \mathbb{R}^*$, so $f(r)$ is constant. But then $f$ is just coming from a ring homomorphism $\mathbb{R} \to \mathbb{R}$ and we are done. The same holds for every polynomial ring over $\mathbb{R}$ (in any number of variables).
With limits we can then build more really unique rings, such as the "axis ring" $\mathbb{R}[X] \times_{\mathbb{R}} \mathbb{R}[Y] \cong \mathbb{R}[X,Y]/\langle XY \rangle$.
Of course every subring of a really unique commutative ring $A$ that contains the copy of $\mathbb{R}$ is also really unique. So for example, $\mathbb{R}[X^2,X^3] \cong \mathbb{R}[X,Y]/\langle X^3-Y^2 \rangle$ is really unique. For every topological space $S$ the ring $C(S)$ of real-valued continuous functions on $S$ is really unique (it is a subring of $\mathbb{R}^S$).
The question is partially motivated by this one, where $A = M_2(\mathbb{R})$ (not commutative) and uniqueness fails. In his answer, Qiaochu also reminded us that $\mathbb{C} \cong \mathbb{R}[X]/\langle X^2+1 \rangle$ is not really unique (in particular, being finitely generated over $\mathbb{R}$ is not sufficient).
Since $\mathbb{C}$ is not really unique, it follows that for a really unique $\mathbb{R} \hookrightarrow A$ the only intermediate field $F$ which is finitely generated over $\mathbb{R}$ must be $\mathbb{R}$ (using Zariski's Lemma). But I don't know what happens when we drop finite generation.
The "universal" example of a non-really unique ring is $\mathbb{R} \otimes \mathbb{R}$ (tensor product over the integers or rationals, doesn't matter). In general, a commutative ring $A$ is really unique if it admits at least one homomorphism from $\mathbb{R}$ and every homomorphism $\mathbb{R} \otimes \mathbb{R} \to A$ factors over the codiagonal $\mathbb{R} \otimes \mathbb{R} \to \mathbb{R}$. But this just a reformulation of the definition.
If a full classification is not possible, answers with interesting examples are also appreciated. We may also restrict to finitely generated $\mathbb{R}$-algebras if necessary.
Here is the proof that $\mathbb{R}$ is really unique. I will write it down here so that we don't need to switch between tabs, but also since I suspect that the method may be generalized. So let $f : \mathbb{R} \to \mathbb{R}$ be a ring homomorphism. Clearly, $f(q)=q$ for all $q \in \mathbb{Q}$, and since $x \geq 0 \iff x \text{ is a square}$ we see that $f$ preserves the order. Also, $f$ is injective since $\mathbb{R}$ is a field, and hence reflects the order. Now $f(r)$ is the supremum of all rational numbers $q$ with $q \leq f(r)$, i.e. $f(q) \leq f(r)$. This is equivalent to $q \leq r$, so the supremum is $r$. $\checkmark$
