It's well-known that sums of independent Cauchy are Cauchy. Indeed, for $Y_1 \sim \text{Cauchy}(m_1, s_1)$ and $Y_2 \sim \text{Cauchy}(m_2, s_2)$, we have characteristic function \begin{align*} \varphi_{Y_1 + Y_2}(t) = Ee^{i(t Y_1 + t Y_2)} = Ee^{it Y_1}Ee^{it Y_2} = e^{m_1 t - s_1|t|}e^{m_2 t - s_2|t|} = e^{i(m_1 + m_2)t - (s_1 + s_2)|t|} \end{align*} and so $Y_1 + Y_2 \sim \text{Cauchy}(m_1 + m_2, s_1 + s_2)$.
Consider now \begin{align*} Z_1 = m_1 + Y + \epsilon_1 \\ Z_2 = m_2 + Y + \epsilon_2 \end{align*} where $Y, \epsilon_1, \epsilon_2$ are independent with $Y \sim \text{Cauchy}(0, s)$, $\epsilon_1, \epsilon_2 \sim \text{Cauchy}(0, 1)$. The characteristic function for $(Z_1, Z_2)$ is \begin{align*} Ee^{i(t_1Z_1 + t_2Z_2)} = e^{i(m_1 t_1 + m_2 t_2)}Ee^{i(t_1+t_2)Y}Ee^{it_1 \epsilon_1}Ee^{it_1 \epsilon_2} = e^{i(m_1 t_1 + m_2 t_2) - s|t_1 + t_2| -|t_1|-|t_2|} \end{align*} This definition does not match the definition for a multivariate Cauchy $Z \sim \text{Cauchy}(m, S)$: \begin{align*} Ee^{it^\intercal Z} = e^{it^\intercal m - \sqrt{t^\intercal S t}} \end{align*} Then, what is the distribution of $(Z_1, Z_2)$? Can the density for this be expressed in closed-form?