I might be wrong, but it looks like the answer is negative.

Let $X$ be the unit circle, that we identify with $[-\pi, \pi]$ with the endpoints glued together. Write $a_+ = \max\{a, 0\}$ and $a_- = \max\{-a, 0\}$. Consider a deterministic process $X_1(t)$ that moves to the right with velocity $1 + 2 (\sin x)_+^{1/2}$. Its generator is $$ \Omega_1 f(x) = (1 + 2 (\sin x)_+^{1/2}) f'(x) , $$ with domain $C^1(X)$.

Edit: In other words, $X_1(t)$ satisfies the differential equation $$ X_1'(t) = (1 + 2 (\sin X_1(t))_+^{1/2}) . $$ By Picard–Lindelöf theorem, this equation induces a continuous flow $\phi_t(x)$ on $(-\pi, 0) \cup (0, \pi)$. Now suppose that $X_1(t)$ is a solution with $X_1(0) = 0$ (such a solution exists by Peano's theorem). Then $X_1'(0) = 1$, and hence $X_1(t) > 0$ for small $t > 0$. Therefore, $X_1(t) = \phi_{t-s}(X_1(s))$ for small $t, s > 0$. It follows that $X_1(t) = \phi_t(0^+)$ for small $t > 0$, and so the solution $X_1(t)$ with starting point $X_1(0) = 0$ is (locally) unique. A similar argument shows uniqueness of solutions with $X_1(0) = \pi$. It follows that the flow can be extended to all of $X$ in a unique continuous fashion. (This certainly follows from some general theorems, too.)

Now consider a similar deterministic process $X_2(t)$ that moves to the left with velocity $-1 - 2 (\sin x)_-^{1/2}$. It is generated by $$ \Omega_2 f(x) = -(1 + 2 (\sin x)_-^{1/2}) f'(x) , $$ again with domain $C^1(X)$. The average of these two operators is $$ \Omega = \tfrac12 (\Omega_1 + \Omega_2) f(x) = ((\sin x)_+^{1/2} - (\sin x)_-^{1/2}) f'(x) $$ for $f$ in $C^1(X)$. However, this operator does not generate a $C_0$ semigroup on $C(X)$.

Indeed: suppose, contrary to our claim, that it does. Then the corresponding Markov process $X(t)$ necessarily satisfies the differential equation $$ X'(t) = ((\sin X(t))_+^{1/2} - (\sin X(t))_-^{1/2}) $$ as long as $X'(t) \in (-\pi, 0) \cup (0, \pi)$. When the starting point goes to $0^+$ or $0^-$, the solutions converge to two (non-unique) solutions with the starting point $0$ (they behave as $\pm \tfrac14 t^2$ as $t \to 0^+$), and so it is easy to see that $e^{t \Omega} f$ can fail to be continuous at $0$, contrary to our assumption.

I might be wrong, but it looks like the answer is negative.

Let $X$ be the unit circle, that we identify with $[-\pi, \pi]$ with the endpoints glued together. Write $a_+ = \max\{a, 0\}$ and $a_- = \max\{-a, 0\}$. Consider a deterministic process $X_1(t)$ that moves to the right with velocity $1 + 2 (\sin x)_+^{1/2}$. Its generator is $$ \Omega_1 f(x) = (1 + 2 (\sin x)_+^{1/2}) f'(x) , $$ with domain $C^1(X)$.

Edit: In other words, $X_1(t)$ satisfies the differential equation $$ X_1'(t) = (1 + 2 (\sin X_1(t))_+^{1/2}) . $$ This is an autonomous first-order ODE, with a unique solution $$X_1(t) = F(F^{-1}(X_1(0)) + t),$$ where $F$ is the indefinite integral of $1 / (1 + 2 (\sin x)_+^{1/2})$. Clearly, $X_1(t)$ depends continuously on $t$ and the initial value $X_1(0)$, and hence $\Omega_1$ indeed generates a $C_0$ semigroup on $C(X)$. (This certainly follows from some general theorems, too.)

Now consider a similar deterministic process $X_2(t)$ that moves to the left with velocity $-1 - 2 (\sin x)_-^{1/2}$. It is generated by $$ \Omega_2 f(x) = -(1 + 2 (\sin x)_-^{1/2}) f'(x) , $$ again with domain $C^1(X)$. The average of these two operators is $$ \Omega = \tfrac12 (\Omega_1 + \Omega_2) f(x) = ((\sin x)_+^{1/2} - (\sin x)_-^{1/2}) f'(x) $$ for $f$ in $C^1(X)$. However, this operator does not generate a $C_0$ semigroup on $C(X)$.

Indeed: suppose, contrary to our claim, that it does. Then the corresponding Markov process $X(t)$ necessarily satisfies the differential equation $$ X'(t) = ((\sin X(t))_+^{1/2} - (\sin X(t))_-^{1/2}) $$ as long as $X'(t) \in (-\pi, 0) \cup (0, \pi)$. When the starting point goes to $0^+$ or $0^-$, the solutions converge to two (non-unique) solutions with the starting point $0$ (they behave as $\pm \tfrac14 t^2$ as $t \to 0^+$), and so it is easy to see that $e^{t \Omega} f$ can fail to be continuous at $0$, contrary to our assumption.

I might be wrong, but it looks like the answer is negative.

Let $X$ be the unit circle, that we identify with $[-\pi, \pi]$ with the endpoints glued together. Write $a_+ = \max\{a, 0\}$ and $a_- = \max\{-a, 0\}$. Consider a deterministic process that moves to the right with velocity $1 + (\sin x)_+^{1/2}$. Its generator is $$ \Omega_1 f(x) = (1 + 2 (\sin x)_+^{1/2}) f'(x) , $$ with domain $C^1(X)$. Now consider a similar deterministic process that moves to the left with velocity $-1 - (\sin x)_-^{1/2}$. It is generated by $$ \Omega_2 f(x) = -(1 + 2 (\sin x)_-^{1/2}) f'(x) , $$ again with domain $C^1(X)$. The average of these two operators is $$ \Omega = \tfrac12 (\Omega_1 + \Omega_2) f(x) = ((\sin x)_+^{1/2} - (\sin x)_-^{1/2}) f'(x) $$ for $f$ in $C^1(X)$. However, this operator does not generate a $C_0$ semigroup on $C(X)$. Indeed: suppose, contrary to our claim, that it does. Then the corresponding Markov process $X(t)$ necessarily satisfies the differential equation $$ X'(t) = ((\sin X(t))_+^{1/2} - (\sin X(t))_-^{1/2}) $$ as long as $X'(t) \in (-\pi, 0) \cup (0, \pi)$. When the starting point goes to $0^+$ or $0^-$, the solutions converge to two (non-unique) solutions with the starting point $0$ (they behave as $\pm \tfrac14 t^2$ as $t \to 0^+$), and so it is easy to see that $e^{t \Omega} f$ can fail to be continuous at $0$, contrary to our assumption.

I might be wrong, but it looks like the answer is negative.

Let $X$ be the unit circle, that we identify with $[-\pi, \pi]$ with the endpoints glued together. Write $a_+ = \max\{a, 0\}$ and $a_- = \max\{-a, 0\}$. Consider a deterministic process $X_1(t)$ that moves to the right with velocity $1 + 2 (\sin x)_+^{1/2}$. Its generator is $$ \Omega_1 f(x) = (1 + 2 (\sin x)_+^{1/2}) f'(x) , $$ with domain $C^1(X)$.

Edit: In other words, $X_1(t)$ satisfies the differential equation $$ X_1'(t) = (1 + 2 (\sin X_1(t))_+^{1/2}) . $$ By Picard–Lindelöf theorem, this equation induces a continuous flow $\phi_t(x)$ on $(-\pi, 0) \cup (0, \pi)$. Now suppose that $X_1(t)$ is a solution with $X_1(0) = 0$ (such a solution exists by Peano's theorem). Then $X_1'(0) = 1$, and hence $X_1(t) > 0$ for small $t > 0$. Therefore, $X_1(t) = \phi_{t-s}(X_1(s))$ for small $t, s > 0$. It follows that $X_1(t) = \phi_t(0^+)$ for small $t > 0$, and so the solution $X_1(t)$ with starting point $X_1(0) = 0$ is (locally) unique. A similar argument shows uniqueness of solutions with $X_1(0) = \pi$. It follows that the flow can be extended to all of $X$ in a unique continuous fashion. (This certainly follows from some general theorems, too.)

Now consider a similar deterministic process $X_2(t)$ that moves to the left with velocity $-1 - 2 (\sin x)_-^{1/2}$. It is generated by $$ \Omega_2 f(x) = -(1 + 2 (\sin x)_-^{1/2}) f'(x) , $$ again with domain $C^1(X)$. The average of these two operators is $$ \Omega = \tfrac12 (\Omega_1 + \Omega_2) f(x) = ((\sin x)_+^{1/2} - (\sin x)_-^{1/2}) f'(x) $$ for $f$ in $C^1(X)$. However, this operator does not generate a $C_0$ semigroup on $C(X)$. 

Indeed: suppose, contrary to our claim, that it does. Then the corresponding Markov process $X(t)$ necessarily satisfies the differential equation $$ X'(t) = ((\sin X(t))_+^{1/2} - (\sin X(t))_-^{1/2}) $$ as long as $X'(t) \in (-\pi, 0) \cup (0, \pi)$. When the starting point goes to $0^+$ or $0^-$, the solutions converge to two (non-unique) solutions with the starting point $0$ (they behave as $\pm \tfrac14 t^2$ as $t \to 0^+$), and so it is easy to see that $e^{t \Omega} f$ can fail to be continuous at $0$, contrary to our assumption.