Define quadratic functions $g_1(X)$ and $g_2(X)$ such that \begin{align}\mathcal{E}_1 &= \{ X \mid g_1(X) \le 0\}\\\mathcal{E}_2 &= \{ X \mid g_2(X) \le 0\}\end{align}
It follows from the definition that you can find a point in $\mathcal{E}_1 \setminus \mathcal{E}_2$ by solving
$$\begin{align} \min &\quad -g_2(X)\\\text{s.t.} &\quad g_1(X) \le 0.\end{align}$$
Informally, you find the point in $\mathcal{E}_1$ with the maximum (scaled) distance from the center of $\mathcal{E}_2$. If the optimal value of $-g_2(X) < 0$, then this point is not in $\mathcal{E}_2$, as required.
The problem above is not obviously convex: if $g_1(X)$ is convex, then the objective $-g_1(X)$ is typically not convex.
However, in this special case (and under some technical assumptions) the S-lemma can be used to reformulate this problem as a convex problem!
For details, see Theorem 2.2 in the survey by Pólik and Terlaky, for example. I personally find the lecture notes of lecture 12 provided here very helpful.
Edit: As pointed out by C Marius and Mark L. Stone, the approach above does not actually result in a point $X \in \mathcal{E}_1 \setminus \mathcal{E}_2$. Instead, it answers whether such a point exists. I looked further into this, and it turns out you can actually recover a point $X$ after solving the reformulated problem.
In this paper, Tuy and Tuan present the S-lemma as a strong duality theorem for specific non-convex problems, including the one above. In this specific case, we have (ignoring some technical details):
$$\inf_{X\in \mathbb{R}^n} \sup_{\lambda\ge 0} \{-g_2(X) + \lambda g_1(X)\} = \sup_{\lambda\ge 0} \inf_{X\in \mathbb{R}^n} \{-g_2(X) + \lambda g_1(X)\}.$$
The left-hand side is the primal problem, which is equivalent to the optimization problem stated at the top ($g_1(X) \le 0$ is optimal, because otherwise $\lambda \rightarrow \infty$ would send the objective to $+\infty$). The right-hand side is the dual problem, which is equivalent to the convex reformulation mentioned above.
After solving the dual problem (the convex reformulation), we obtain an optimal dual value $\bar{\lambda}$. You then recover a primal solution in a similar way as for linear programming: by using the optimality conditions (Tuy and Tuan, Theorem 3):
$$-\nabla g_2(X) + \bar{\lambda} \nabla g_1(X) = 0 \wedge \bar{\lambda} g_1(X) = 0 \wedge g_1(X) \le 0.$$
Example: Consider two ellipses, given by $A_1 = I$, $B_1 = (-1, 0)^\top$, $C_1 = 0$ (blue circle), and $A_2 = I$, $B_1 = 0$, $C_2 = -1$ (red circle)
The dual is given by
$$\begin{align} \max_{V, \lambda} &\quad V\\
\text{s.t.} &\quad \begin{bmatrix} \lambda A_1 - A_2 & \lambda B_1 - B_2 \\ (\lambda B_1 - B_2)^\top & \lambda C_1 - C_2 - V \end{bmatrix} \succeq 0 \\
~&~\lambda \ge 0.\end{align}$$
Solving this SDP yields the solution $V = -3$ and $\bar{\lambda} = 2$. As $V < 0$, there exists an $X \in \mathcal{E}_1 \setminus \mathcal{E}_2$.
If we use the first of the optimality conditions, we have
$$-\nabla g_2(X) + \bar{\lambda} \nabla g_1(X) = 2 (\bar{\lambda} A_1 - A_2)X + 2(\bar{\lambda} B_1 - B_2) = 2IX - (-4, 0)^\top = 0,$$
which solves for $\bar{X} = (2,0)^\top$. It is easy to check that the other conditions are also satisfied, which means that $\bar{X}$ is a primal optimal solution. Indeed, $\bar{X} \in \mathcal{E}_1 \setminus \mathcal{E}_2$.
In the picture, the yellow circle corresponds to the ellipse $g_2(X) = -V$. This ellipse intersects $\mathcal{E}_1$ at the maximum (weighted) distance from the center of $\mathcal{E}_2$. Their intersection is exactly the point $\bar{X}$.
I have omitted various technical details, for which I refer to the references.