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added more detail, made the notation a bit more consistent with the links, and directly addressed the question asked by the OP

The reason a Langevin diffusion leaves $\nu(x)=e^{-U(x)}$ invariant is because it is symmetric or reversible with respect to $\nu$. In comparison to general diffusion processes, the ergodic properties of symmetric diffusions are not only easier to analyze, but it is often possible to obtain an explicit formula for the stationary density of a symmetric diffusion. This symmetry property is a bit more transparent when the Langevin equation is written as $$ dX_t = \nabla \log \nu(X_t) dt + \sqrt{2} dW_t \;. \tag{$\star$} $$ The symmetry property can then be easily verified by checking that the infinitesimal generator of the Langevin diffusion is symmetric in an $L^2$-inner product weighted by $\nu$. This property of the Langevin diffusion is analogous to the detailed balance condition of a Markov chain. It is also the basis of standard MCMC methods based on Langevin diffusions. In fact, many random-walk based MCMC methods (like random walk Metropolis) with target probability density proportional to $\nu(x)$, can weakly approximate the corresponding Langevin diffusion ($\star$). This is remarkable because, e.g., the random walk Metropolis algorithm does not involve the gradient of $U$.

To read more about this connection, see the introduction and Theorem 5.2 of this paper.