More generally, we have the following result: If $R$ is an artinian ring and $R \to S$ is a surjective ring homomorphism, then also $R^* \to S^*$ is surjective.

Proof: We may assume that $R$ is local (otherwise $R$ is a direct product of such rings and $R \to S$ decomposes into a product of such homomorphisms, etc.), and also $S \neq 0$. Then $S = R/p$ for some nilpotent ideal $p$. Since $1+p$ consists of units, it is even true that every preimage of a unit in $S$ is also a unit in $R$.

In the special case $R=\mathbb{Z}/mn$ this gives the proofs above using the Chinese Remainder Theorem.

More generally, we have the following result: If $R$ is an artinian commutative ring and $R \to S$ is a surjective ring homomorphism, then also $R^* \to S^*$ is surjective.

Proof: We may assume that $R$ is local (otherwise $R$ is a direct product of such rings and $R \to S$ decomposes into a product of such homomorphisms, etc.), and also $S \neq 0$. Then $S = R/p$ for some nilpotent ideal $p$. Since $1+p$ consists of units, it is even true that every preimage of a unit in $S$ is also a unit in $R$.

In the special case $R=\mathbb{Z}/mn$ this gives the proofs above using the Chinese Remainder Theorem.