Actually, that uniqueness theorem is not properly used in the example you mention. As it stands in your initial statement, the uniqueness property holds when $\Omega$ is an open, connected and **bounded** subset of $\mathbb R^n$ and $\varphi$ is continuous in $\overline{\Omega}= \Omega \cup \partial \Omega$ and it is $C^2(\Omega)$ satisfying Poisson's equation $\Delta \varphi = \rho$ in $\Omega$ itself. 

(The proof of uniqueness is a trivial consequence of a celebrated theorem on harmonic functions $\phi$ in $\Omega$, i.e., functions verifying $\Delta \phi =0$ in $\Omega$, which are  continuous in $\overline{\Omega}$. That theorem  establishes that, if $\Omega$ is open and it closure is compact, $\max_{\overline{\Omega}} |\phi|$ is reached in a point of $\partial \Omega$. Thinking of $\phi$ as the difference of two solutions of Poisson's equation filling the same boundary conditions, the uniqueness property easily arises.)

When $\Omega$ is **not** bounded, as in the mentioned example, where $\Omega = \{(x,y,z) \in {\mathbb R}^3\:|\: z>0\}$, one must add further requirements on the behaviour  of $\varphi$ for $||(x,y, z)|| \to +\infty$, and there are a number of possibilities.
 
In this case in $\Omega$ the charge is $q$ (that is the $\rho$ distribution you mention) with distance $d$ from $\partial \Omega$ and $\varphi=0$ on $\partial \Omega$ because $\partial \Omega$ is a grounded conducting plane. The value of $\varphi$ on $\partial \Omega$ is constant, we are free to assume that it is $0$. *The true boundary condition here is that $\varphi$ attains on $\partial \Omega$ the same value it attains for   $||(x,y, z)|| \to +\infty$.*

Let us pass to consider the situation where two charges stay at the reciprocal distance $2d$ along the $z$ axis, focussing on what happens in $\overline{\Omega}$ (not outside it) regarding charge distributions and boundary conditions of $\varphi$. 

 The $\rho$ distribution  in the half space $\Omega = \{(x,y,z) \in {\mathbb R}^3\:|\: z>0\}$ is the same as in the previous case: There is the charge $q$ at distance $d$ from the plane at $z=0$.

Also the boundary conditions of $\varphi$ on $\partial \Omega$ and for  $||(x,y, z)|| \to +\infty$ are the same as for the other case: The plane at $z=0$ is equipotential in view of the symmetry of the problem and the value of $\varphi$ thereon is the same as the value of $\varphi$ for  $||(x,y, z)|| \to +\infty$. 

Therfore, applying the uniqueness property in $\overline{\Omega}$, we are committed to conclude that  the potential $\varphi$ in the region $\Omega \cup \partial \Omega$ is the same in both cases.