$\newcommand{\R}{\mathbb R}$
Take $\Omega\subset \R^d$ bounded,convex, and smooth.
Consider the optimal transport problem with cost $c(x,y)=|x-y|^2$, leading to the quadratic Wassersein distance. Are there known optimal (?) conditions on the probability measures $\mu,\nu\in \mathcal P(\Omega)$ guaranteeing uniqueness (up to additive constants) of the Kantorovich potentials $\varphi$ from $\mu$ to $\nu$?

One standard answer is to require that $\mu$ be absolutely continuous (w.r.t) the $d$-dimensional Lebesgue measure $\mathcal L^d_{\Omega}$.
Indeed a Kantorovich potential $\varphi$ is always Lipschitz (at least in bounded domains) hence differentiable Lebesgue-almost everywhere (Rademacher's theorem), and therefore also $\mu$-almost everywhere.
Then one can prove that $y-x=\nabla\varphi(x)$ for any $(x,y)$ in the support of any optimal plan, which gives uniqueness of both the plans and potentials (and as a byproduct that $T(x)=x+\nabla\varphi(x)$ is the optimal Monge's map s.t. $T_\#\mu=\nu$).
This is well explained e.g. in Filippo Santambrogio's book [1, theorem 1.17 pp. 15].

However, for my sepcific purpose I'm interested in a case where $\mu=f_\mu \mathcal L^d|_\Omega+g_\mu\mathcal L^{d-1}|_{\partial\Omega}$ and $\nu=f_\nu \mathcal L^d|_\Omega+g_\nu\mathcal L^{d-1}|_{\partial\Omega}$ typically have a nice smooth absolutely continuous part in the interior $\Omega$ but also a $(d-1)$-dimensional component on the boundary ($\mathcal L^k$ denotes here the $k$-dimensional Lebesgue measure).
The above sufficient condition obviously fails, and I am left wondering what can be said? I still expect somehow that the Kantorovich potential (which always exists) is unique, but I have no real clue how to proceed and I was hoping someone out there would have a "black-box reference" that I could use before reinventing the wheel.