What is known is explained in C. Albanese, S. Lawi, Laplace transform of an integrated gometric Brownian motion, MPRF 11 (2005), 677-724, in particular in the paragraph of the Introduction beginning by A separate class of models...
First of all, note that the natural filtration of $(X_t)_{t \geq 0}$ equals the natural filtration of $(W_t)_{t \geq 0}$; this follows directly from the relation
$$X_t = \exp \left( \left[ \mu - \frac{\sigma^2}{2} \right] t + \sigma W_t \right).$$
The Markov property of $(X_t)_{t \geq 0}$ can be proved as follows: Fix some bounded Borel-measurable function $f$ and $s \leq t$. For brevity, set $c := \mu - \frac{\sigma^2}{2}$. Then
$$\begin{align*} \mathbb{E}(f(X_t) \mid \mathcal{F}_s) &= \mathbb{E} \bigg( f \left[ e^{ct} e^{\sigma(W_t-W_s)} e^{\sigma W_s} \right] \mid \mathcal{F}_s \bigg). \end{align*}$$
Since $W_t-W_s$ is independent from $\mathcal{F}_s = \sigma(W_r; r \leq s)$ and $W_s$ is $\mathcal{F}_s$-measurable, we get
$$\mathbb{E}(f(X_t) \mid \mathcal{F}_s) = g(e^{ct} e^{\sigma W_s}) \tag{1}$$
where
$$g(y) := \mathbb{E} \left( f(y e^{\sigma (W_t-W_s)}) \right).$$
Since the right-hand side of $(1)$ is $X_s$-measurable, the tower property yields
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E} \bigg[ \mathbb{E}(f(X_t) \mid \mathcal{F}_s) \mid X_s \bigg] \stackrel{(1)}{=} g(e^{ct} e^{\sigma W_s}). \tag{2}$$
Combining $(1)$ and $(2)$ we find
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E}(f(X_t) \mid \mathcal{F}_s).$$
Best Answer
Hint
Using the fact that $W_t-W_s$ is independent of $\mathcal F_s$ yields
$$\mathbb E[\alpha e^{\sigma W_t-\sigma ^2t/2}\mid \mathcal F_s]=\mathbb E[e^{\sigma (W_t-W_s)}]e^{\alpha W_s+\sigma ^2t/2}.$$