$\newcommand\om\omega\newcommand\Om\Omega\newcommand\ep\epsilon$Let $(P_t)_{t\in T}$ be an exponential family over a separable complete metric space $(X,d)$, where $T$ is an open subset of $\mathbb R^k$ and $t=(t_1,\dots,t_k)$ is a natural parameter. So, for all $t\in T$, the density $p_t$ of $P_t$ with respect to Borel measure $\mu$ on $X$ is given by the formula $$p_t(x)=c(t)\exp\{t\cdot U(x)\}\tag1$$ for all $x\in X$, where $U\colon X\to\mathbb R^k$ is a Borel-measurable function, $\cdot$ denotes the dot product on $\mathbb R^k$, and $$c(t):=1\Big/\int_X\mu(dx)\exp\{t\cdot U(x)\}$$ is the normalizing factor. Assume that, for some (and hence for all) $x_0\in X$, the "first moment" $$m_t(x_0):=\int_X P_t(dx)d(x_0,x)=\int_X \mu(dx)p_t(x)d(x_0,x) $$ is finite for all $t\in T$.
Then the parameterization map $U\ni t\mapsto P_t$ is continuous with respect to the Wasserstein-1 distance $W_1$.
Indeed, by the penultimate paragraph of the introduction section, if suffices to show that for each $t\in T$ $$\int_X \mu(dx)p_s(x)f(x)\to \int_X \mu(dx)p_t(x)f(x)$$ as $s\to t$, where $f\colon X\to\mathbb R$ such that $\sup_{x\in X}|f(x)|/(1+d(x_0,x))<\infty$ (for some (and hence for all) $x_0\in X$). This follows immediately by the uniform integrability.
In turn, in view of (1), the uniform integrability here follows by the following argument: Take any $t=(t_1,\dots,t_k)$ in the open set $T$. Take any real $\ep>0$ such that $t+2\ep\om\in T$ for all $\om=(\om_1,\dots,\om_k)\in\Om:=\{-1,1\}^k$. Then for any $s=(s_1,\dots,s_k)\in T$ such that $\max_{j=1}^k|s_j- t_j|\le\ep$ we have \begin{align}\exp\{s\cdot u\}/\exp\{t\cdot u\}&=\exp\{(s-t)\cdot u\} \\ &\le\max_{\om\in\Om}\exp\{\ep\om\cdot u\} \\ &=\exp\Big\{\ep\sum_{j=1}^k|u_j|\Big\} \\ &<<\exp\Big\{2\ep\sum_{j=1}^k|u_j|\Big\} \\ &=\max_{\om\in\Om}\exp\Big\{2\ep\om\cdot u\Big\} \\ &\le\sum_{\om\in\Om}\exp\Big\{2\ep\om\cdot u\Big\}; \end{align} the first inequality in the above display follows because the function $\exp$ is convex and the condition $\max_{j=1}^k|s_j- t_j|<\ep$ implies that the point $s-t$ is in the convex hull of the set $\{\ep\om\colon\om\in\Om\}$; here $a<<b$ mean that $a=o(b)$ as $\sum_{j=1}^k|u_j|\to\infty$. So, uniformly over all $s\in T$ such that $\max_{j=1}^k|s_j- t_j|\le\ep$ we have \begin{align}\exp\{s\cdot u\}<<\sum_{\om\in\Om}\exp\{( t+2\ep\om)\cdot u\}, \end{align} and $t+2\ep\om\in T$ for all $\ep=(\ep_1,\dots,\ep_k)\in\Om:=\{-1,1\}^k$. Thus, the uniform integrability follows. $\Box$