Perhaps instead of finite sets one should work with effective divisors, i.e., formal linear combibinations of the form $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bp}{\boldsymbol{p}}$
$$ D= \sum_{\bp\in\bR^N} m_D(\bp) \bp, $$
where $m_D:\bR^N\to\bZ_{\geq 0}$ is a function with finite support. Denote by $\newcommand{\Div}{\mathrm{Div}}$ $\Div_{\geq 0}(\bR^N)$ the space of effective divisors in $\bR^N$. $\Div_{\ge 0}$ has an obvious semigroup structure. Think of the center of mass as a map $\newcommand{\eC}{\mathscr{C}}$
$$ \eC: \Div_{\geq 0}(\bR^N)\to \Div_{\geq 0}(\bR^N), $$
with the following properties.
1. For any divisor $D$, the support of the center of mass $\eC(D)$ consists of a single point.
2. For any divisors $D_1,D_2\in\Div_{\geq 0}(\bR^N)$ we have
$$ \eC(D_1+D_1) =\eC\Bigl(\; \eC(D_1)+\eC(D_2)\; \Bigr).$$
3. For any point $\bp\in \bR^N$ and any $m\in\bZ_{\geq 0}$
$$\eC(m \delta_{\bp})=m\delta_{\bp}, $$
where $\delta_{\bp}$ denotes the Dirac divisor of mass $1$ supported at $\bp$.
4. $\eC$ is continuous with respect to the obvious topology on $\Div_{\geq 0}(\bR^N)$, where supports of divisors converge in the Hausdorff metric and in the limit the total mass is conserved.
Claim. I believe that these properties uniquely determine $\eC$.
Weaker Claim. The map $\eC$ is uniquely determined if besides 1,...4 we assume the following additional condition.
5. If the support of $D$ is contained in an affine subspace $V\subset \bR^N$, then the support of $\eC(D)$ is contained in the same affine subspace.
Edit 1. Strengthened Condition $3$. The function
$$ \Div_{\geq 0}(\bR^N)\ni D=\sum_{\bp} m(\bp)\delta_{\bp}\mapsto \eC_0(D):=m(D)\delta_{C(D)} \in \Div_{\geq 0}(\bR^N)\tag{$\ast$}, $$
where
$$ m(D) :=\left(\sum_{\bp} m(\bp)\right) ,\;\; c(D)= \frac{1}{m(D)}\sum_{\bp} m(\bp) \bp, $$
satisfies all the above $5$ conditions. The Weak Claim states that this is the only function satisfying these conditions.
Edit 2. I took into account Deanne's Deane's comments and I was able to prove the following version of the above claim.
PROPOSITION. There exists exactly one map $\eC:\Div_{\geq 0}(\bR^N)\to \Div_{\geq 0}(\bR^N)$ satisfying the following conditions.
(Localization.) The support of $\eC(D)$ consists of a single point.
(Conservation of mass.)
$$ m\bigl( D\bigr)=m\bigl(\;\eC(D)\;\bigr),\;\;\forall D\in \Div_{\geq 0}$$
(Normalization.)$\newcommand{\bq}{\boldsymbol{q}}$ $$\eC(m\delta_{\bp}) =m\delta_{\bp},\;\; \eC(\delta_{\bp}+\delta_{\bq})=2\delta_{\frac{1}{2}(\bp+\bq)}. $$
(Additivity.)
$$\eC(D_1+D_2)=\eC\bigl(\;\eC(D_1)+\eC(D_2)\;\bigr),\;\;\forall D_1,D_2\in\Div_{\geq 0}. $$
Edit 3. The proof of the above proposition does not use the linear structure of $\bR^N$. It uses only the metric structure, and more precisely the fact that any two different points in $\bR^N$ determine a unique geodesic. The normalization condition should be rewritten as
$$\eC(\delta_{\bp}+\delta_{\bq})= 2 \delta_{c(\bp,\bq)}, $$
where $c(\bp,\bq)$ denotes the midpoint of the geodesic arc $[\bp,\bq]$. In the above proposition we can then replace $\bR^N$ with the hyperbolic space $\newcommand{\bH}{\boldsymbol{H}}$ $\bH^N$ and we can conclude that on a hyperbolic space there exists at most one notion of center of mass, i.e., a map $\eC:\Div_{\geq 0}(\bH^N)\to \Div_{\geq 0}(\bH^N)$ satisfying the four conditions in the Proposition.