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. $mathbb{Z} to mathbb{Z}/pmathbb{Z}$ 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.
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