There is a simple and reasonably general sufficient criterion for a ring surjection $f : R \to S$ to induce a surjection $f^\times : R^\times \to S^\times$ on unit groups (apologies for bumping an old post, but none of the other answers seemed to have this simple line of reasoning).

**Proposition:** Let $f : R \twoheadrightarrow S$. If $\ker f$ is contained in all but finitely many maximal ideals of $R$, then $f^\times$ is surjective.

Proof: Write $I := \ker f$, and $\text{mSpec}(R) \setminus V(I) = \{m_1,...,m_n\}$. Then $\{I, m_1,...,m_n\}$ are pairwise comaximal. Pick $v \in S^\times$, and write $v = f(u)$ for some $u \in R$ (notice $u \not \in m$, for any $m \in \text{mSpec}(R) \cap V(I)$). By Chinese Remainder, there exists $a \in R$ with $a \equiv 0 \pmod{I}$, $a \equiv 1-u \pmod{m_i}$ for $i = 1,...,n$. Then $u + a \in R^\times$, and $f(u+a) = f(u) = v$.

This immediately yields that if $R$ is semilocal (has only finitely many maximal ideals), then every surjection out of $R$ induces a surjection on units. This generalizes the case where $R$ is Artinian (or finite). The case that $I$ is contained in the Jacobson radical of $R$ can also be recovered via the reduction:

**Proposition:** Let $\overline{R} := R/\text{rad}(R)$ (where $\text{rad}(R)$ is the Jacobson radical). Then $f^\times : R^\times \to (R/I)^\times$ is surjective iff $\overline{f}^\times : \overline{R}^\times \to (\overline{R}/\overline{I})^\times$ is surjective.

This also yields the semilocal case, since then $\overline{R}$ is a finite product of fields. Concerning the limitations of the first proposition: although the condition that $I$ avoids only *finitely* many maximal ideals seems strong, it is in a sense sharp: e.g. $\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ for $p$ prime, $p > 3$ does not induce a surjection on units. One final remark that may be of interest:

**Proposition:** If $R = \bigoplus_{i=0}^\infty R_i$ is $\mathbb{N}$-graded and $I \subseteq R_+$ is a homogeneous prime concentrated in positive degree, then $R^\times \to (R/I)^\times$ is surjective.