The divergence theorem in $\mathbb{R}^3$ says that the integral of the divergence of a vector field $V$ over a solid $\Omega$ in $\mathbb{R}^3$ equals the flux through the surface of $\Omega$ denoted by $\partial \Omega$: $$ \int_\Omega \operatorname{div} V \, dx = \int_{\partial \Omega} (V \cdot n) \, dA $$
Does this uniquely define the divergence operator?
Yes it does. Let us formalize the result a bit: Let $P: C^1(\mathbb{R}^n,\mathbb{R}^n)\rightarrow C^0(\mathbb{R}^n)$ be some operator with the property: $\forall V\in C^1(\mathbb{R}^n,\mathbb{R}^n)$, $\forall \Omega\subset\mathbb{R}^n$ with $\Omega$ bounded and smooth boundary, we have $$\int_\Omega PV\, dx = \int_{\partial\Omega} \langle V,N\rangle\, dA$$ and $N$ is the outer unit normal of $\Omega$. Then $PV=\operatorname{div} V$. Let us proof this:
Since any Ball $B_R(x)\subset\mathbb{R}^n$ is bounded and has smooth boundary, we have by divergence theorem $$\int_{B_R(x)}PV - \operatorname{div}V\, dx = \int_{\partial B_R(x)} \langle V,N\rangle\, dA-\int_{\partial B_R(x)} \langle V,N\rangle\, dA=0.$$ Therefore $$0=\frac{1}{\mathcal{L}^n(B_R(x))} \int_{B_R(x)}PV - \operatorname{div}V\, dx\rightarrow PV(x) - \operatorname{div}V(x)\mbox{ for }R\downarrow 0$$ Hence $PV=\operatorname{div}V$. The convergence is true, because for continuous functions, every point is a Lebesgue point.
Or speaking from a physics viewpoint, you have the equality $$\int_\Omega PV\, dx = \int_\Omega \operatorname{div} V\, dx$$ for all test volumina $\Omega$, hence the integrands have to be the same.