Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution of $Y_0\in\hat X$, then $Y$ will have a continuous modification which necessarily satisfies $Y_t\not\in\hat X\setminus X$ for all positive times (almost surely). In your example with $Y$ being a Brownian motion and $\hat X=\mathbb{R}^3\setminus\{0\}$ then this process corresponds to adding back the origin.

This is actually a special case of a more general method called *Ray-Knight compactification*, which applies to right-continuous Markov processes taking values in a Suslin space. Ray-Knight compactification does not always lead to processes which are Feller though, as they can have branch points in the extended state space. The type of processes obtained by this method are called *Ray processes*. See the 1975 paper by Getoor & Sharpe detailing Ray-Knight compactification, or, any reasonably comprehensive textbook on Markov processes should describe this method. In your question the conditions that $X$ is locally compact, $T_tC_0\subseteq C_b$ and $Y$ is continuous are enough to ensure that we obtain a Feller process so, in particular, there will be no branch points in the extended domain. One point before continuing; I'm assuming, as is standard in the definition of Feller processes, that the space $X$ has a countable base (locally compact, Hausdorff with a countable base, aka an *lccb* space).

Constructing the extension is a rather natural application of the Gelfand-Naimark theorem, although showing that the resulting process is Feller will still require some work. Let $\mathcal{A}$ be the smallest closed $C^*$-subalgebra of $C_b(X)$ containing $C_0(X)$ and closed under application of $T_t$. We can define $\hat X$ to be its spectrum (the set of nonzero $C^*$ homomorphisms $\mathcal{A}\to\mathbb{C}$), under the weak topology. Then, we identify $X$ as an (open, dense) subset of the (locally compact) space $\hat X$ in the same way as in the construction of the Stone-Chech compactification. By the Gelfand-Naimark theorem, every $f\in\mathcal{A}$ extends uniquely to an $\hat f\in C_0(\hat X)$, giving an isometry $\mathcal{A}\to C_0(\hat X)$, $f\mapsto\hat f$. Then, $T_t$ extends to a map $\hat T_t\colon C_0(\hat X)\to C_0(\hat X)$, $\hat T_t\hat f=\hat{T_tf}$. This is automatically a Markov transition function, but we can show that it is also Feller in your case. It is a bit of work, and I'll break it up into smaller statements.

1) $T_tC_b(X)\subseteq C_b(X)$.

Any nonnegative $f\in C_b(X)$ is a pointwise limit of an increasing sequence $f_n\in C_0(X)$ and, therefore $T_tf=\lim_{n\to\infty}T_tf_n$ is also a limit of an increasing sequence in $C_0(X)$, so is lower semicontinuous. Applying the same statement to $\Vert f\Vert - f$ implies that $f$ is also upper semi-continuous, so is continuous.

2) If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)_+\to0$ in the compact-open topology.

(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.)
It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int_0^{1/m}T_sf\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So,
$$
T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon.
$$
on $K$. Letting $t$ decrease to zero gives $\limsup_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$.

3) For any $f\in\mathcal{A}$ and $t_n\to0$, $T_{t_n}f\to f$ *uniformly* as $n\to\infty$.

This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$,
$$
\begin{align}
T_tf(x)=\mathbb{E}_x[f(Y_t)]&=\mathbb{E}_x\left[T_{(t-\tau)_+}f(Y_{\tau\wedge t})\right]\\\\
&\le\sup_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)_+.
\end{align}
$$
The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$.

Now let $f_n$ be the sequence converging uniformly on compacts to $f$ constructed in the proof of 2. Then, we have $f_n\to f$ uniformly on $X\setminus K$ as $n\to\infty$ so, in fact, $f_n$ tends uniformly to $f$. Taking the limits $t\to0$ and $n\to\infty$ in
$$
\Vert T_tf-f\Vert\le\Vert T_tf_n-f_n\Vert+2\Vert f_n-f\Vert
$$
shows that $\Vert T_tf-f\Vert\to0$.

This almost shows that $\hat T$ is Feller, just the following technical lemma is left.

4) $\hat X$ is an lccb space.

As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base.

So, $\hat T_t$ is a Feller transition function on $\hat X$. Therefore, any Markov process $\hat Y$ w.r.t. this transition function has a right-continuous modification. We can also show that, regardless of the initial distribution, $\hat Y$ will be continuous and $\hat Y_t\not\in\hat X\setminus X$ for all times $t > 0$ (almost-surely). For any time $s > 0$ at which $Y_s\in X$ then the statement of the question implies that $\hat Y$ has a continuous modification lying in $X$ at all times $t\ge s$ (which is unique). So, it is enough to show that $\mathbb{P}(\hat Y_s\in X)=1$ for each time $s > 0$. But, this is standard (see Prop. 4.6 (iii) from the paper of Getoor & Sharpe).