$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}$Yes, the minimizer $\mu$ is absolutely continuous (w.r. to the Lebesgue measure $|\cdot|$).
Indeed, you showed that
\begin{equation*}
F(\rho_n)\to m:=\inf_\rho F(\rho) \tag{-1}
\end{equation*}
and
\begin{equation*}
\mu_{\rho_n}\to\mu \tag{-0.5}
\end{equation*}
weakly for some sequence $(\rho_n)$ of probability densities and some probability measure $\mu$, where
\begin{equation*}
F(\rho):=\int\rho\ln\rho\,dx+W_2^2(\rho,\eta) \tag{0}
\end{equation*}
and $\mu_\rho(dx):=\rho(x)\,dx$.
Take any set $E\subseteq\R$ with $|E|=0$. We have to show that then $\mu(E)=0$.
Take any real $\ep>0$. By the regularity of the Lebesgue measure, there is an open set $G_\ep\subset\R$ such that
\begin{equation*}
\text{$E\subseteq G_\ep$ and $|G_\ep|<\ep$.} \tag{0.5}
\end{equation*}
By (-0.5) and the Portmanteau theorem,
\begin{equation*}
\mu(G_\ep)\le\liminf_n\mu_{\rho_n}(G_\ep). \tag{1}
\end{equation*}
Next, for each real $a>1$,
\begin{equation*}
\mu_{\rho_n}(G_\ep)=K_n+L_n, \tag{2}
\end{equation*}
where
\begin{equation*}
K_n:=\int_{G_\ep\cap[\rho_n\le a]}\rho_n\,dx,\quad L_n:=\int_{G_\ep\cap[\rho_n>a]}\rho_n\,dx,
\end{equation*}
$[\rho_n\le a]:=\rho_n^{-1}((-\infty,a])$, $[\rho_n>a]:=\rho_n^{-1}((a,\infty))$.
Further,
\begin{equation*}
K_n\le a|G_\ep|<a\ep \tag{3}
\end{equation*}
by (0.5), and
\begin{equation*}
L_n\le \int_{[\rho_n>a]}\rho_n\,dx
\le\frac1{\ln a}\int \rho_n\ln\rho_n\,dx\le \frac{m+1}{\ln a} \tag{4}
\end{equation*}
for all large enough $n$, by (-1) and (0).
By (0.5), (1), (2), (3), (4),
\begin{equation*}
\mu(E)\le\mu(G_\ep)\le a\ep+\frac{m+1}{\ln a},
\end{equation*}
for all real $\ep>0$ and all real $a>1$. Letting now $\ep\downarrow0$ and then $a\to\infty$, we get $\mu(E)=0$, as desired.
$\newcommand{\R}{\mathbb R}\newcommand{\H}{\mathcal H}$Building up on previous comments, and slighlty elaborating. The answer to your question, as is, is NO. Recall that convergence in the Wasserstein distance $W_2$ is equivalent to weak-* convergence (duality with continuous bounded functions) and convergence of the second moments (or equivalently, with all continuous functions growing at most quadratically at infinity). As a consequence, in order for your compactness to hold you would need to control the second moments in sublevelsets. However, the entropy is translation-invariant, so mass can escape at infinity in a sublevelset $\{\H\leq C\}$. More precisely, for a counterexample take as in @Kostya_I 's comment
any measure $\rho$ with finite entropy $\H(\rho)=C<\infty$, take any $x_n\in\R^d$ such that $|x_n|\to\infty$, and consider the translated sequence $\rho_n=\rho(\cdot-x_n)$. Then obviously $\H(\rho_n)=\H(\rho)=C$ for any $n$, but $\rho_n$ fails to converge in the Wasserstein distance (since for example $\rho_n$ converges weakly to zero when tested with respect to any compactly supported function).
Since the problem here is only due mass escaping at infinity, for your desired compactness to hold you only need a little bit of confinement.
For example it would suffice to control the second moment $\mathfrak m_2(\rho)=\int|x|^2\rho(dx)$, for example by adding a potential energy
$$
\mathcal F(\rho)=\H(\rho)+\int V(x)\rho(dx)
$$
for some potential $V(x)\gtrsim C|x|^2$ growing at least quadratically at infinity.
For this you can use the handy Carleman estimate
$$
\H(\rho)\geq -C_\alpha(1+\mathfrak{m}_2(\rho))^\alpha,
\qquad
\alpha\in \big(d/(d+2),1\big).
$$
Check the original JKO paper [1], proof of proposition 4.1 (eqs. 14 and 15 therein).
Actually you can replace the potential energy $\int V\rho$ by anything controlling $\mathfrak m_2(\rho)^{\frac{d}{d+2}}$ in order to retrive control over the moments from energy bounds $\mathcal F(\rho)\leq C\Rightarrow \mathfrak m_2(\rho)\leq M(C)$.
Connecting with @pseudocydonia 's answer above: another way to interpret the "potential energy above" would be to consider instead the relative entropy $\H_\mu(\rho)=\int \frac{\rho}{\mu}\log\left(\frac{\rho}{\mu}\right)d\mu$ for a log-quadratic reference measure $\mu\sim e^{-V}$ with $V\gtrsim C|x|^2$.
Side note: in your setting you also have a slightly worse problem than what you may suspect. When computed relatively to the Lebesgue measure on (sufficiently wide) unbounded sets the entropy is actually not bounded from below on $\mathcal P_{ac}^2$.
[1] Jordan, R., Kinderlehrer, D., & Otto, F. (1998). The variational formulation of the Fokker--Planck equation. SIAM journal on mathematical analysis, 29(1), 1-17.
Best Answer
$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.
Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the probability measures with densities $\rho_n$ converge in total variation and hence in $\tau$ to the probability measure with density $\rho$.
On the other hand (assuming that your $\log$ is $\ln$), $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim -n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$