[All thanks and credit to saz in the comments.]
The answer is yes. Let's proceed step by step.
For $n\in\mathbb N^*$ and $t=(t_1,\cdots,t_n)$, define
$$P^n_t:F\mapsto\big(x\mapsto\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]\big)\text.$$
Suppose that $F:E^n\to\mathbb R$ is of the form $x_1,\cdots,x_n\mapsto g_1(x_1)\cdots g_n(x_n)$ with $g_i\in\mathcal C$.
Then $P^n_tF\in\mathcal C$.
The proof is by induction on $n$. The case $n=1$ is the Feller property. Now if the statement holds for $n-1$, then $\phi:x\mapsto\mathbb E_x[g_2(X_{t_2-t_1}),\cdots,g_n(X_{t_n-t_1})]$ must be in $\mathcal C$, and so would $x\mapsto g_1(x)\phi(x)$. Since $\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]=\mathbb E_x[g_1(X_{t_1})\phi(X_{t_1})]$ by the Markov property, we can apply the Feller property and conclude.
Suppose that $F:E^n\to\mathbb R$ is continuous and vanishes at infinity (with respect to the product topology).
Then $P^n_tF$ is continuous bounded.
Boundedness is clear. The continuity can be shown by an approximation argument. Let $A$ be the algebra of finite sums of functions of the form $x_1,\cdots,x_n\mapsto g_1(x_1)\cdots g_n(x_n)$ with $g_i$ continuous vanishing at infinity. According to the locally compact Stone-Weierstrass theorem, there exists a sequence $(G_k)_{k\geq0}$ of elements of $A$ such that $G_k\to F$ in the uniform sense. But this implies that $P^n_tG_k\to P^n_tF$ in the uniform sense, and since the former is continuous according to the previous claim, then the latter must be as well.
Suppose in addition that $E$ is second countable.
Then for any $F:E^n\to\mathbb R$ continuous bounded, $x\mapsto\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]$ is continuous bounded.
Again, boundedness is obvious. To show continuity, we fix $x\in E$ and show that $P^n_tF$ is continuous at $x$.
Fix $\varepsilon>0$. The first step is to find a suitable cutoff function to ignore what happens at infinity. Because $E$ is second countable locally compact, it is $\sigma$-compact. In particular, there exists a compact $K\subset E$ such that
$$\mathbb P_x(X_{t_i}\notin K\text{ for some }1\leq i\leq n)\leq\frac\varepsilon{3|F|_\infty}\text.$$
Because $E$ is locally compact Haussdorff, there exists a continuous function $\chi:E\to[0,1]$ with compact support such that $\chi\geq1_K$; it is a generalisation of Urysohn's lemma (for some reason, the generalisation appears on the French Wikipedia). By definition,
$$ \mathbb E_x[\chi(X_{t_1})\cdots\chi(X_{t_n})]
\geq \mathbb P_x[X_{t_1}\in K\text,\cdots\text,X_{t_n}\in K]
\geq 1-\frac\varepsilon{3|F|_\infty}\text.$$
But using the above result, we see that $y\mapsto \mathbb E_y[\chi(X_{t_1})\cdots\chi(X_{t_n})]$ is continuous; in particular, there exists a neighbourhood $\mathcal U$ of $x$ such that
$$ \mathbb E_y[\chi(X_{t_1})\cdots\chi(X_{t_n})] \geq 1-\frac\varepsilon{2|F|_\infty}$$
for any $y\in\mathcal U$.
Now that we found our cutoff function, it is time to conclude. For any $y\in\mathcal U$, we have
$$ P^n_tF(y)
= \mathbb E_y[F(X_{t_1},\cdots,X_{t_n})\chi(X_{t_1})\cdots\chi(X_{t_n})]
+ \mathbb E_y\big[F(X_{t_1},\cdots,X_{t_n})(1-\chi(X_{t_1})\cdots\chi(X_{t_n}))\big]
= (1)_y + (2)_y\text.$$
By construction, $|(2)_y|$ is at most $\varepsilon/2$. Moreover, according to the previous statement, $(1)_y$ is continuous with respect to $y$, and
$$ |(1)_x- P^n_tF(x)| \leq |F|_\infty\mathbb E_x[1-\chi(X_{t_1})\cdots\chi(X_{t_n})]
\leq \frac\varepsilon3\text. $$
In particular, we have $|(1)_y-P^n_tF(x)|<\varepsilon/2$ for all $y$ in a neighbourhood $\mathcal V$ of $x$, which proves that for any $y\in\mathcal U\cap\mathcal V$,
$$|P^n_tF(y)-P^n_tF(x)|<\varepsilon\text.$$
This concludes the proof.
I am now more convinced that the proof is quite immediate. I will supply the necessary details in order to conclude what I have already alluded to in my finalizing remarks above.
Let $\alpha \geq 0$ and $f_\alpha(x) :=\int_0^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Since $P_tf\in{C}_0(\mathbb{R}^d)$ by the definition above (2), as $f \in {C}_0(\mathbb{R}^d)$ and since every function continuous and tending to $0$ at $\infty$ is bounded, one has $\alpha e^{-\alpha t}||P_tf(\cdot)||_\infty<C\cdot\alpha e^{-\alpha t}$ for some constant $C>0$. Thus one may apply dominated convergence to obtain
$\lim_{x\rightarrow0}f_{\alpha}(x)=\int_0^{\infty}\alpha e^{-\alpha t}\lim_{x\rightarrow0}P_tf(x)dt=0$
showing $f_\alpha \in {C}_0(\mathbb{R}^d)$. Now let $f\geq0$ and define $f_{\alpha,h}(x)=\int_h^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Setting $f(t,x):=\alpha e^{-\alpha t}(P_tf)(x)$, one has that $\lim_{x\rightarrow0}f(t,x)=0$ $\forall$ $t\in(h,\infty)$ by the definition above point (2) and $f(t,x)$ is measurable in $t$ $\forall$ $x\in\mathbb{R}^d$. Thus one has
$\lim_{x\rightarrow\infty}f_{\alpha,h}(x)=\lim_{x\rightarrow\infty}\int_h^{\infty}\alpha e^{-\alpha t}(P_tf)(x)dt=\int_h^{\infty}\lim_{x\rightarrow\infty}\big(\alpha e^{-\alpha t}(P_tf)(x)\big)dt$
by Lebesgues theorem as the integrand is bounded in $x$, showing that $f_{\alpha,h}\in{C}_0(\mathbb{R}^d)$. Now by the definition point (3)
$\lim_{t\downarrow0}(P_tf_{\alpha,h})(x)=\lim_{t\downarrow0}P_t\int_{h}^\infty\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}(x)$,
finally showing the claim.
Best Answer
I don't see how to give a short proof of the stated identity (... perhaps I'm missing something). I will use the following general statement on conditional expectations which is a consequence of Lévy's convergence theorem
Fix $t>0$. By the continuity of $U_{\alpha}f$ and the càdlàg property of the sample paths, we have $$U_{\alpha} f(X_{(T+t)-}) = \lim_{n \to \infty} U_{\alpha} f(X_{T_n+t})$$ for any sequence of stopping times $T_n$ with $T_n \uparrow T$. Applying the above lemma with $$Y := U_{\alpha} f(X_{(T+t)-}) \qquad Y_n := U_{\alpha} f(X_{T_n+t}) \qquad \mathcal{G}_n := \mathcal{F}_{T_n}$$ we find that
\begin{align*} \mathbb{E}(U_{\alpha} f(X_{(T+t)-}) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}) &= \lim_{n \to \infty} \mathbb{E}(U_{\alpha} f(X_{T_n+t}) \mid \mathcal{F}_{T_n}). \end{align*}
It now follows from the strong Markov property that
\begin{align*} \mathbb{E}(U_{\alpha} f(X_{(T+t)-}) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}) &= \lim_{n \to \infty} P_t(U_{\alpha}f)(X_{T_n}) \tag{1} \end{align*}
If we let $t \downarrow 0$, then it follows from the fact that $U_{\alpha}f$ is bounded and continuous and that $X$ has càdlàg sample paths that the left-hand side of $(1)$ converges to
$$\mathbb{E}(U_{\alpha}f(X_T) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}).$$
It remains to show that the right-hand side of $(1)$ converges to $\lim_{n \to \infty} U_{\alpha} f(X_{T_n})$ as $t \to 0$. To this end, we note that, by the Feller property, $$\|P_t (U_{\alpha}f)- (U_{\alpha} f)\|_{\infty} \xrightarrow[]{t \to 0} 0,$$
and so
\begin{align*} \limsup_{t \to 0} \left| \lim_{n \to \infty} P_t(U_{\alpha} f)(X_{T_n}) - \lim_{n \to \infty} U_{\alpha} f(X_{T_n}) \right| &= \limsup_{t \to 0} \lim_{n \to \infty} |P_t (U_{\alpha} f)(X_{T_n})-(U_{\alpha}f)(X_{T_n})| \\ &\leq \limsup_{t \to 0} \|P_t(U_\alpha f) - U_{\alpha} f\|_{\infty} =0. \end{align*}