$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$The answer to Question 1 becomes yes if, as suggested by the OP in a personal message, we use the following definition of the entropy instead of the one in the OP: 
$$\H(\rho):=\int_{\R^d}U(\rho(x))\,dx,$$
where $U(r):=r\ln r+1-r$ for real $r>0$, with $U(0):=1$, so that $U$ is a **nonnegative** convex function on $[0,\infty)$. 

Indeed, suppose that a sequence of probability measures $\mu_n$ with densities $\rho_n$ converges in $\tau$ to a probability measure $\mu$ with density $\rho$. For real $\si>0$, let $\ga_\si$ be the Gaussian measure over $\R^d$ with mean $0$ and covariance matrix $\si I_d$, where $I_d$ is the identity matrix. Let $f_\si$ be the density of $\ga_\si$. 

Then for each $\si>0$ the density $f_\si *\rho_n$ of the probability measure $\ga_\si *\mu_n$ converges pointwise to the density $f_\si *\rho$ of the probability measure $\ga_\si *\mu$ (this follows because $f_\si$ is a bounded continuous function). So, by the Fatou lemma, 
$$\liminf_n\H(f_\si *\rho_n)\ge \H(f_\si *\rho). \tag{1}\label{1}$$

By Jensen's inequality applied to the convex function $U$, 
$$\H(\rho_n)\ge\H(f_\si *\rho_n). \tag{2}\label{2}$$ 

Also, $f_\si *\rho\to\rho$ almost everywhere as $\si\downarrow0$. So, again by the Fatou lemma, 
$$\liminf_{\si\downarrow0}\H(f_\si *\rho)\ge \H(\rho). \tag{3}\label{3}$$

It follows by \eqref{2},\eqref{1}, and \eqref{3} that $\liminf_n\H(\rho_n)\ge \H(\rho)$. So, $\H$ is l.s.c. in $\tau$. $\quad\Box$


  [1]: https://mathoverflow.net/questions/457609/are-the-sublevel-sets-of-boltzmann-entropy-compact-in-wasserstein-distance#comment1185275_457609