$\DeclareMathOperator\Irr{Irr}$You can see this from the fact that for abelian groups, irreducible characters form a $\mathbb{C}$-basis of the space of functions from $G$ to $\mathbb{C}$. These functions correspond bijectively to linear transformations from the group algebra $\mathbb{C}G$ to $\mathbb{C}$, and $\Irr(G)$ is also a basis for the space of such linear transformations. Therefore, $\chi(\sum_{x\in X}x)=\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)=\chi(\sum_{y\in Y}y)$ for each $\chi\in \Irr(G)$ implies that $f(\sum_{x\in X}x)=f(\sum_{y\in X}y)$ for every linear transformation $f:\mathbb{C}G\rightarrow \mathbb{C}$. This happens if and only if $\sum_{x\in X}x=\sum_{y\in Y}y$.

There is another way to see this. In general, for split finite dimensional algebra $A$ over a field $k$, the common zero set of irreducible characters $\Irr_k(A)$ is $J(A)+[A,A]$, where $J(A)$ is the Jacobson radical of $A$ and $[A,A]$ is the commutator subalgebra of $A$ generated by elements of the form $[a,b]=ab-ba$. 

If $G$ is a finite abelian group, then the group algebra $\mathbb{C}G$ is finite dimensional, split, semisimple and commutative. Hence, both $J(A)$ and $[A,A]$ are trivial, and the above theorem says that the common zero set of irreducible complex characters is trivial. In particular, if two elements $a$, $b$ of the group algebra satisfy $\chi(a)=\chi(b)$ for each $\chi\in \Irr(G)$, then $a-b$ is in the common zero set, which is $\{0\}$.