Expanding upon @User1865345's and @Ute's answers, here is another way of understanding product measures on $\mathcal B(\mathbb R^d)$, the Borel $\sigma$-algebra on $\mathbb R^d$, via their associated (multivariate) CDFs:
Recall that for each (multivariate) CDF $F: \mathbb R^d \to \mathbb R$ there is exactly one probability measure $\mathbb P_F$ on $\mathcal B(\mathbb R^d)$ with CDF $F$, i.e.,
$$
\mathbb P_F((-\infty, t_1] \times \ldots \times (-\infty, t_d]) = F(t_1, \ldots, t_d) \;\forall\,t_1, \ldots, t_d \in \mathbb R
$$
holds for exactly one $\mathbb P_F$.
The CDF of the product measure $\mathbb P_{F_1} \times \cdots \times \mathbb P_{F_d}$ is, $\forall\,t_1, \ldots, t_d \in \mathbb R$, given by
\begin{align}
F_{\mathbb P_{F_1} \times \cdots \times \mathbb P_{F_d}}(t_1, \ldots t_d)
&= (\mathbb P_{F_1} \times \cdots \times \mathbb P_{F_d})((-\infty, t_1] \times \ldots \times (-\infty, t_d]) \\
&= \mathbb P_{F_1}((-\infty, t_1]) \cdot \ldots \cdot \mathbb P_{F_1}((-\infty, t_d]) \\
&= F_1(t_1) \cdot \ldots \cdot F_d(t_d),
\end{align}
i.e., by the product of the corresponding marginal (univariate) CDFs.
Now, letting $X_1 \sim F_1, \ldots, X_d \sim F_d$, we find that $(X_1, \ldots, X_d) \sim \mathbb P_{F_1} \times \cdots \times \mathbb P_{F_d}$ iff the joint CDF of $(X_1, \ldots, X_d)$ factorizes into the marginal CDFs (i.e., iff $X_1, \ldots, X_d$ are independent).
In the example in the arXiv paper you link to, $P_x$ is a probability measure on $\mathcal B(\mathbb R^p)$ and $P_\epsilon$ is a probability measure on $\mathcal B(\mathbb R)$.
Thus, $(x_i, \epsilon_i) \sim P_x \times P_\epsilon$ in equation $(1)$ just means that the joint CDF $F_{x_i, \epsilon_i} \equiv F_{P_x \times P_\epsilon}$ of $(x_i, \epsilon_i)$ is the product of the marginal CDFs $F_{x_i} \equiv F_{P_x}$ and $F_{\epsilon_i} \equiv F_{P_\epsilon}$ of $x_i$ and $\epsilon_i$, respectively:
$$
F_{x_i, \epsilon_i}(t_1, \ldots, t_p, t_{p+1}) = F_{x_i}(t_1, \ldots, t_p) \cdot F_{\epsilon_i}(t_{p+1}) \;\forall\,t_1, \ldots, t_{p+1} \in \mathbb R.
$$