$\newcommand{\si}{\sigma}\newcommand{\W}{\mathcal W}\newcommand{\R}{\mathbb R} $Yes, the map
\begin{equation*}
\mu \mapsto J(\mu):=\int_\R T^\si_\mu(x+y)\exp(-y^2/2\si^2)\,dy \tag{10}\label{10}
\end{equation*}
is continuous on the 2-Wasserstein space, say $\W_2$.
Indeed, let $\mu_n\to\mu$ in $\W_2$ (as $n\to\infty$). Suppose that $X_n\sim\mu_n$ and $X\sim\mu$. Then $X_n\to X$ in distribution and $EX_n^2\to EX^2$. Letting $Z$ be a standard normal random variable independent of the $X_n$'s and $X$, we get $X_n+\si Z\to X+\si Z$ in distribution, so that
\begin{equation*}
F_{(\mu_n)_\si}\to F_{\mu_\si} \tag{20}\label{20}
\end{equation*}
pointwise.
Also, the function $F_{\nu_\si}$ is continuous and strictly increasing, so that the inverse function $F^{-1}_{\nu_\si}$ is continuous (and strictly increasing). So, by \eqref{20},
\begin{equation*}
T_{\mu_n}^\si\to T_\mu^\si \tag{30}\label{30}
\end{equation*}
pointwise.
By Markov's inequality, $F_{\nu_\si}(z)\le C/z^2$ for real $z<0$, where $C:=\int_\R x^2\,\nu(dx)<\infty$, and hence
\begin{equation*}
|F^{-1}_{\nu_\si}(u)|\le\sqrt{C/u} \tag{40}\label{40}
\end{equation*}
if $0<u<F_{\nu_\si}(0)$ (and $F_{\nu_\si}(0)\in(0,1)$).
On the other hand, letting $m_n$ and $m$ denote the medians of $(\mu_n)_\si$ and $\mu_\si$, respectively, and letting $\Phi_\si$ denote the cdf of $N(0,\si^2)$, we see that \eqref{20} implies $m_n\to m$ and hence
\begin{equation*}
F_{(\mu_n)_\si}(z)=\int_\R F_{\mu_n}(z-y)d\Phi_\si(y) \\
\ge\int_{-\infty}^{z-m_n}F_{\mu_n}(m_n)d\Phi_\si(y)
=\frac12\,\Phi_\si(z-m_n)=\exp\left(-\frac{z^2}{2\si^2+o(1)}\right)
\end{equation*}
uniformly in $n$ as $z\to-\infty$. So, by \eqref{40},
\begin{equation*}
|T_{\mu_n}^\si(z)|=|F^{-1}_{\nu_\si}(F_{(\mu_n)_\si}(z))|
\le \exp\frac{z^2}{4\si^2+o(1)} \tag{50}\label{50}
\end{equation*}
uniformly in $n$ as $z\to-\infty$. Similarly, \eqref{50} holds in the right-tail zone, that is, uniformly in $n$ as $z\to\infty$.
Also, in view of \eqref{20} and because $F_{\mu_\si}$ is strictly increasing, we see that
\begin{equation*}
|T_{\mu_n}^\si(z)|=O(1)
\end{equation*}
uniformly in $n$ for $|z|=O(1)$. So, in view of \eqref{50},
\begin{equation*}
|T_{\mu_n}^\si(z)|=O\Big(\exp\frac{z^2}{3\si^2}\Big)
\end{equation*}
uniformly in $n$ and real $z$. So, for each fixed real $x$,
\begin{equation*}
|T^\si_{\mu_n}(x+y)|\exp(-y^2/2\si^2)
=O\Big(\exp-\frac{y^2}{7\si^2}\Big)
\end{equation*}
uniformly in $n$ and real $y$.
Thus, by \eqref{10}, \eqref{30}, and dominated convergence, $J(\mu_n)\to J(\mu)$. $\quad\Box$
