Let me attempt a proof using the group-theoretic formulation. I will use the additive notation for the group operation.
The proof is by induction on $n=|G|$, with the base being trivial. Let $n=p^rm$ for some prime $p$ with $\gcd(p,m)=1$.
Consider $G' = pG$. Our goal is to reduce the problem for $G$ to its instance for $G'$. Let $R_j:=G' + jm$. Then $\{R_0,R_1,\ldots, R_{p-1}\}$ is a partition of $G$ into $G'$-cosets.
Let $G_i=d_iG$ be a subgroup of $G$ with $d_i | n$. If $p|d_i$ then $G_i \subseteq G'$ and every $G_i$ coset belongs to some $R_j$. In this case
we say that $G_i$ is of the first kind. Otherwise, $d_i | m$, and translating $G_i$ by $m$ does not change $G_i$. We say that such $G_i$ is of the second kind.
Let $S = \cup_{i=1}^k (G_i+x_i)$ be the union under consideration, let $S_1$ be the union of cosets of the first kind among cosets comprising $S$, and $S_2$ -- of the second. Let $T_j=S_1 \cap R_j$ for $0\leq j \leq p-1$. Then $T_j$ is actually union of some of our cosets, and the sets $T_0,T_1,\ldots,T_{p-1}$ are disjoint. Let $T_j'=T_j-jm$: we shift all the cosets in $T_j$ from $R_j$ to $R_0=G'$. Note that $S_2-jm=S_2$ and therefore the intersection of $T_j$ and $S_2$ shifts with $T_j$. In particular, $|T_j'-S_2|=|T_j-S_2|$.
The set $S'=S_2 \cup (\cup_{j=0}^{p-1}T'_j) $ is still a union of cosets of $G_i$'s. We have
$|S'| \leq |S_2|+ \sum_{j=1}^{p-1}|T'_{j}-S_2| = |S_2|+ \sum_{j=1}^{p-1}|T_{j}-S_2|=|S|$.
Thus it suffices to consider $S'$, which means that we may assume that $S_1 \subseteq G'$. Let $G_i'=G_i \cap G'$ and let
$x_i'$ be chosen so that $G_i'+x_i'= (G_i+x_i) \cap G'$. By the induction hypothesis applied to $G'$, we get
$a_1:=| \cup_{i=1}^k G_i' | \leq | \cup_{i=1}^k (G_i' + x_i')|=: b_1$
and also
$a_2:=|\cup_{i: G_i \not \subseteq G'} G_i'| \leq |\cup_{i: G_i \not \subseteq G'} (G_i' + x_i')|=:b_2$,
where in this second inequality we restrict our attention to $G_i$'s of the second kind. The set $S_2$ is the disjoint union of $p$ translates of its intersection with $G'$, which intersection is present on the right side of the inequality directly above. It follows that
$|S|=|S \cap G'|+|S_2 - G'|=b_1 + (p-1)b_2,$
while similarly we have
$|\cup_{i=1}^k G_i|=a_1 + (p-1)a_2$.
It follows that $|S| \geq |\cup_{i=1}^k G_i|$, as desired.
Finally, let me note that the inequality does not hold for non-cyclic groups. Already for $G = \mathbf{Z}_2 \times \mathbf{Z}_2$ the union of three distinct subgroups of $G$ of size $2$ is $G$, while it is possible to choose their cosets with the union of size $3$.
