The answer is yes: every infinite finitely generated group has a subset which is equidecomposable to a proper subset of itself.

This follows from a theorem of Brandon Seward: Every finitely generated infinite
group $G$ admits a translation-like action by the group $\mathbb{Z}$ of integers. https://arxiv.org/abs/1104.1231

What this means is that there exists a free action of $\mathbb{Z}$ on $G$ such that each element of $\mathbb{Z}$ acts as an element of the wobbling group of $G$. That is, there exists a bijection $f:G\rightarrow G$ having no finite orbits, along with a finite partition $G=\bigsqcup_{i=1}^nX_i$ and finitely many group elements $a_1,\dots , a_n\in G$ such that $f(x)=a_ix$ for all $x\in X_i$. If we take $T$ to be a subset of $G$ containing exactly one point from each orbit, and define $A$ to be the set $A = \bigsqcup _{n\geq 0} f^n(T)$, then $f(A)$ is a proper subset of $A$, and the partition $A=\bigsqcup _{i=1}^n A_i$, where $A_i := A\cap X_i$, along with the group elements $a_1,\dots , a_n$, witness that $A$ is equidecomposable with $f(A)$.

**Edit:** I realized there is also a simple direct argument which also shows that $n$ is bounded above by the minimum size of a finite subset $S$ of $G$ which generates an infinite subsemigroup of $G$. Consider the (left) Cayley graph of this subsemigroup $H$ with respect to its generating set $S$, i.e., with vertex set $H$ and with a directed edge from $h$ to $sh$ for each $s\in S$ and $h\in H$. This graph is infinite and locally finite, so by Kőnig's lemma it contains an infinite geodesic ray, say with vertices $h_0,h_1,\dots$. Let $A := \{ h_0,h_1,h_2,\dots \}$ and let $B=A \setminus \{ h_0 \}$. For each $n\geq 0$ there is some $s_n\in S$ such that $h_{n+1}=s_nh_n$, so $A$ is partitioned into finitely many sets $A_s$, $s\in S$, where $A_s$ consists of those $h_n$ for which $s_n=s$. Then $B$ is partitioned by $sA_s$, $s\in S$, so $A$ and $B$ are equidecomposable.