It this case, if for example $m_1\neq m_2, \sigma_1\sigma_2\rho\neq 0$, no, because the drift is not a function of $X_t + Y_t$.
In a general situation, when $X,Y$ are two non independant Markov chains, $X+Y$ is not.
To understand this with a more simple example, in a discret time setting, take for example $(X_n)$ be a simple random walk on $\mathbb Z$, and take $Y_n = n1_{X_1 = 1}$ (is is a Markov process, because is $Y_{n+1} = Y_n + 1_{X_1 = 1}$).
Then, $S = X+Y$ is not a Markov process: the law of $S_n|S_{n-1}$ depends on $X_1 = S_1$.
We have to assume that the underlying probability space is complete; otherwise the assertion might fail.
So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)$$ is continuous for all $\omega \in \tilde{\Omega} := \Omega \backslash N$. Now
$$\tilde{X}_t(\omega) := \begin{cases} X_t(\omega), & \omega \in \tilde{\Omega}, \\ 0, & \omega \in N \end{cases}$$
defines a stochastic process on $\Omega$ with continuous sample paths, and therefore
$$\sup_{t \in [0,1]} \tilde{X}_t = \sup_{t \in [0,1] \cap \mathbb{Q}} \tilde{X}_t$$
is measurable as countable supremum of measurable random variables. On the other hand, we have
$$\tilde{S}_t(\omega) = \sup_{t \in [0,1]} \tilde{X}_t(\omega) = \sup_{t \in [0,1]} X_t(\omega)= S_t(\omega) \quad \text{for all $\omega \in \tilde{\Omega} = \Omega \backslash N$}$$
and so
$$\{S_t \in B\} = \left( \{\tilde{S}_t \in B \} \cap N^c \right) \cup \bigg( \{S_t \in B \} \cap N \bigg)$$
for any Borel set $B$. Since $N \in \mathcal{A}$ and $\tilde{S}_t$ is measurable, we know that
$$\left( \{\tilde{S}_t \in B \} \cap N^c \right) \in \mathcal{A}.$$
Moreover,
$$\left\{ S_t \in B \right\} \cap N \subseteq N$$
and since the probability space is complete, this implies
$$\left\{ S_t \in B \right\} \cap N \in \mathcal{A}.$$
Combining both considerations proves $\{S_t \in B\} \in \mathcal{A}$, and this proves the measurability of $S_t$.
Remark More generally, the following statement holds true in complete probability spaces:
Let $(\Omega,\mathcal{A},\mathbb{P})$ and $(E,\mathcal{B},\mathbb{Q})$ be two measure spaces and assume that $(\Omega,\mathcal{A},\mathbb{P})$ is complete. Let $X, Y: \Omega \to E$ be two mappings. If $X$ is measurable and $X=Y$ almost surely, then $Y$ is measurable.
Best Answer
By applying Ito's formulq we get $e^{\gamma t}X_t$ is a local martingale, so it is a supermartingale for it is positive, and so it is for $X_t$. Then by some classical argument(for example an exercise in Chapter 1 of Brownian motion and stochastic calculus) we know any positive supermartingale converge almost surely. Denote its limit by $X_{\infty}\geq 0$, Fatou's lemma gives
$$E[X_{\infty}]\leq\liminf_{t\rightarrow\infty}E[X_t]=0$$
which allows to show $X_{\infty}=0$ a.s.