**Addendum.** I am raising this addendum at the top because it is more relevant to the question than what I have written before.

We denote by $R^{\times}$ the unit group of a unital ring $R$.

**Claim 1.** Let $R$ be a commutative and unital ring. Then the following are equivalent:

This claim was inspired by Theorem 5.1 of [5], a preprint which cannot be more on topic!

Let us define the Bass stable rank of an associative and unital ring $R$.
We call $\mathbf{r} \in R^n$ a *unimodular row of $R$* if the components of $\mathbf{r}$ generate $R$. We denote by $\operatorname{Um}_n(R)$ the set of unimodular rows of size $n$.
We say that $n > 0$ lies in the *stable range of $R$* if for every $(r_1, \dots, r_{n + 1}) \in \operatorname{Um}_{n + 1}(R)$, there is $r \in R$ such that $(a_1 + ra_{n + 1}, \dots, a_n + r a_{n + 1}) \in \operatorname{Um}_n(R)$. It is well-known and easily checked that the stable range has no gap, i.e., if $n$ lies in the stable range of $R$, then so does $m$ for every $m > n$. It is also well-known that $\operatorname{sr}(R/\mathcal{J}(R)) = \operatorname{sr}(R)$ where $\mathcal{J}(R)$ denotes the *Jacobson radical of $R$*, that is, the intersection of the maximal ideals of $R$.

The Bass *stable rank $\operatorname{sr}(R)$ of $R$*, or simply stable rank of $R$, is the smallest element in the stable range of $R$.

The proof of Claim $1$ is actually straighforward.

*Proof of Claim $1$.* Assume that $R^{\times} \rightarrow (R/I)^{\times}$ is surjective for every ideal $I$ of $R$. Let $(a_1, a_2) \in \operatorname{Um}_2(R)$. Then $a_1 + I$ is unit of $R/I$ with $I = Ra_2$. Therefore we can find $u \in R^{\times}$ such that $a_1 + I = u + I$, i.e., there is $r \in R$ verifying $a_1 + r a_2 = u$. As a result $\operatorname{sr}(R) = 1$.
Let us assume now that $\operatorname{sr}(R) = 1$. Let $I$ be an ideal of $R$ and let $a + I \in (R/I)^{\times}$. Then we can find $b \in I$ such that $(a, b) \in \operatorname{Um}_2(R)$. Because $R$ is of stable rank $1$, there is $r \in R$ such that $u \Doteq a + rb$ is a unit of $R$.

Semilocal rings, and Artinian rings in particular, are known to have stable rank $1$. More generally, so are the rings $R$ such that $R/\mathcal{J}(R)$ has Krull dimension $0$ (Justin Chen's semifields), e.g., von Neumann regular rings.
The reader should be bear in mind that $$\operatorname{sr}(R) \le \dim_{Krull}(R) + 1$$ by the Bass Stable Range Theorem and its generalisation by R. Heitmann [2].

The class of *local-global* rings encompass all the previous classes.

**Definition.** A commutative and unital ring $R$ is *local-global* if every (possibly multi-variate) polynomial $f$ with coefficients in $R$ whose values generate $R$ represents a unit of $R$, i.e., some value of $f$ is a unit.

The following is well-known and immediate.

**Claim 2.** If $R$ is local-global then $\operatorname{sr}(R) = 1$.

*Proof.* Given $(a, b) \in \operatorname{Um}_2(R)$, consider the polynomial $f(X) = a + bX$.

**Former answer.**

I would like to contribute in outlining two natural generalisations of the original question for which the Jacobson radical also comes into play. Eventually, I would like to quote two research results for which OP’s question is essential.

Considering that $R^{\times} = GL_1(R)$, it is natural to ask

When does the surjection $R \twoheadrightarrow R/I$ induce a surjective map $GL_n(R) \rightarrow GL_n(R/I)$?.

Here $I$ is an ideal of $R$ and the second map is the reduction of matrix coefficients modulo $I$. If $I$ is an ideal contained in the Jacobson radical $\text{rad}(R)$ of $R$, then $GL_n(R) \rightarrow GL_n(R/I)$ is surjective [5, Exercise I.1.12.iv].

Considering that $R$ is an $R$-module and that every element in $R^{\times}$ generates $R$, it is natural to ask

For $M$ an $R$-module, when do $R$-generating sets of $M/IM$ lift to $R$-generating sets of $M$?.

This holds if $I \subseteq \text{rad}(R)$ and $M/IM$ is finitely generated over $R$ by Nakayama’s lemma, see also the related concept of superfluous submodule.

I introduce now a result published in 2003 which directly relates to OP's question. Let $R$ be a Dedekind domain which is universal for $GE_2$. The latter property means that in the subgroup $GE_2(R) \subseteq GL_2(R)$ generated by the transvections and the diagonal matrices, the latter generators are only subject to the « obvious » or universal relations (equivalently $K_2(2, R)$ is generated by symbols as a normal subgroup of $St_2(2, R)$ [1, 4]).
Then we have:

If $p$ is a prime element of $R$ such that the natural $R
\twoheadrightarrow R/(p)$ induces a surjective homomorphism between unit groups, then the localization $RS^{-1}$ of $R$ with $S = \{1, p, p^2, \dots\}$ is also universal for $GE_2$.

This was proved in [3] and the result was shown to be sharp in [4].

I can’t resist mentioning a humble result of mine, which I would like to believe is entertaining.

A finitely generated group $G$ of rank $n$ is said to satisfy the *generalised Andrews-Curtis conjecture* if every $n$-tuple of elements which normally generate $G$ can be transitioned to an $n$-tuple of generators of $G$ by means of finitely many $AC$-moves, i.e., applying transvections or replacing one component by a conjugate a finite number of times. (If $G$ is the free group on $n$ generators, this is just the Andrews-Curtis conjecture, still unsettled.) Then the result reads as:

The solvable Baumslag-Solitar group $\langle a, b \,\vert\, aba^{-1} = b^n \rangle$ with $n \ge 1$ satisfies the generalised Andrews-Curtis conjecture if and only if the natural map $R^{\times} \rightarrow (R/(n - 1)R)^{\times}$ is surjective where $R = \mathbf{Z}[1/n]$.

[1] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.

[2] R. Heitmann, "Generating non-Noetherian modules efficiently", 1984.

[3] E. Abe and J. Morita, "Tits systems with affine Weyl groups in Chevalley groups over Dedekind domains", 1988.

[4] H. Yu and S. Chen, "Subrings in quadratic fields which are not $GE_2$", 2003.

[5] C. Weibel, "The K-book", 2013.

[6] J. Chen, "Surjections of unit groups and semi-inverses", 2017.