I assume that $S_n\equiv\{(x_i^{\top},\epsilon_i)\}_{i-1}^n$ and $S_{0,m}\equiv\{(x_{0,i}^{\top},\epsilon_{0,i})\}_{i=1}^m$are independent copies of $(x^{\top},\epsilon)$ with $\mathsf{E}[\epsilon\mid x]=0$ a.s. Then, indeed, a reasonable estimator of $\mathsf{E}y$ is
$$
\hat{\mathsf{E}}y=\hat{\mathsf{E}}x^{\top}\times\hat{\beta},
$$
where
$$
\hat{\mathsf{E}}x=\frac{1}{m}\sum_{i=1}^m x_{0,i} \quad\text{and}\quad \hat{\beta}=\left(\frac{1}{n}\sum_{i=1}^n x_ix_i^{\top}\right)^{-1}\frac{1}{n}\sum_{i=1}^n x_iy_i.
$$
Since $S_n$ and $S_{0,m}$ are independent,
$$
\mathsf{E}[\hat{\mathsf{E}}y]=\mathsf{E}[\hat{\mathsf{E}}x^{\top}]\times \mathsf{E}[\hat{\beta}]=\mathsf{E}x^{\top}\beta,
$$
and
$$
\mathsf{E}[\hat{\mathsf{E}}y]^2=\mathsf{E}\left[\hat{\mathsf{E}}x^{\top}\left(\mathsf{E}\hat{\beta}\hat{\beta}^{\top}\right)\hat{\mathsf{E}}x\right]=\beta^{\top}W_m\beta+\sigma^2\operatorname{tr}(W_mV_n),
$$
where $V_n=\mathsf{E}\left[\sum_{i=1}^n x_ix_i^{\top}\right]^{-1}$ and $W_m\equiv\mathsf{E}[\hat{\mathsf{E}}x\hat{\mathsf{E}}x^{\top}]=\mathsf{E}x\mathsf{E}x^{\top}+m^{-1}\operatorname{Var}(x)$ (we assume that $V_n$ exists and is finite). Consequently,
$$
\bbox[cornsilk,5px]
{
\begin{align}
\operatorname{Var}(\hat{\mathsf{E}}y)&=\mathsf{E}[\hat{\mathsf{E}}y]^2-(\mathsf{E}[\hat{\mathsf{E}}y])^2\\
&=\sigma^2\mathsf{E}x^{\top}V_n\mathsf{E}x+\frac{1}{m}\left(\beta^{\top}\operatorname{Var}(x)\beta+\sigma^2\operatorname{tr}(\operatorname{Var}(x)V_n)\right).
\end{align}
}
$$
It follows that $\operatorname{Var}(\hat{\mathsf{E}}y)\to\sigma^2\mathsf{E}x^{\top}V_n\mathsf{E}x$ as $m\to\infty$.
If you are interested in the conditional variance of $\hat{\mathsf{E}}y$ given $X_n\equiv\{x_i\}_{i=1}^n$, then using the same reasoning one obtains
$$
\bbox[cornsilk,5px]
{
\operatorname{Var}(\hat{\mathsf{E}}y\mid X_n)=\sigma^2\mathsf{E}x^{\top}\tilde{V}_n\mathsf{E}x+\frac{1}{m}\left(\beta^{\top}\operatorname{Var}(x)\beta+\sigma^2\operatorname{tr}(\operatorname{Var}(x)\tilde{V}_n)\right),
}
$$
where $\tilde{V}_n=\left(\sum_{i=1}^n x_ix_i^{\top}\right)^{-1}$. Similarly to the previous case,
$$
\operatorname{Var}(\hat{\mathsf{E}}y\mid X_n)\to \Sigma_n\equiv\sigma^2\mathsf{E}x^{\top}\tilde{V}_n\mathsf{E}x
$$
as $m\to\infty$. Finally, by the SLLN (assuming that $\mathsf{E}\|x\|^2<\infty$ and $\mathsf{E}xx^{\top}$ is positive definite),
$$
n\Sigma_n\to \sigma^2\mathsf{E}x^{\top}\left(\mathsf{E}xx^{\top}\right)^{-1}\mathsf{E}x\quad\text{a.s.}
$$
You may be interested in deriving the asymptotic distribution of $\hat{\mathsf{E}}y$. For each $k\in\mathbb{N}$, we estimate $\hat{\mathsf{E}}y$ using two samples $\{(y_i,x_i^{\top})^{\top}\}_{i=1}^{n_k}$ and $\{x_{0,i}\}_{i=1}^{m_k}$ with $n_k,m_k\to\infty$ as $k\to\infty$. Let $\alpha_k$ be s.t.
$$
\frac{\alpha_k}{\sqrt{m_k}}\to c_1<\infty \quad\text{and}\quad \frac{\alpha_k}{\sqrt{n_k}}\to c_2<\infty.
$$
Then by the WLLN and CLT (assuming that $\mathsf{E}\|x\|^2<\infty$ and $\mathsf{E}xx^{\top}$ is positive definite), using the fact that two samples are independent,
\begin{align}
\alpha_k\left(\hat{\mathsf{E}}y-\mathsf{E}y\right)&=\frac{\alpha_k}{\sqrt{m_k}}\hat{\beta}^{\top}\sqrt{m_k}\left(\hat{\mathsf{E}}x-\mathsf{E}x\right)+\frac{\alpha_k}{\sqrt{n_k}}\hat{\mathsf{E}}x^{\top}\sqrt{n_k}\left(\hat{\beta}-\beta\right) \\
&=c_1\beta^{\top}N_1+c_2\mathsf{E}x^{\top}N_2+o_p(1),
\end{align}
where $N_1\sim \mathcal{N}(0,\operatorname{Var}(x))$, $N_2\sim \mathcal{N}(0,\sigma^2(\mathsf{E}xx^{\top})^{-1})$, and $N_1$ and $N_2$ are independent. Therefore,
$$
\bbox[cornsilk,5px]
{
\alpha_k\left(\hat{\mathsf{E}}y-\mathsf{E}y\right)\xrightarrow{d}\mathcal{N}\left(0,c_1^2\beta^{\top}\operatorname{Var}(x)\beta+c_2^2\sigma^2\mathsf{E}x^{\top}\left(\mathsf{E}xx^{\top}\right)^{-1}\mathsf{E}x\right).
}
$$
