If this SDE is the only one you are concerned about, the answer would definitely be Yes. Just as what you might have been aware, the SDE reads
$$
{\rm d}X_t=\alpha{\rm d}t+{\rm d}\left(\sigma_1W_t^1+\sigma_2W_t^2\right),
$$
where the term in the parenthesis is no more than another Brownian motion, not standard though. Thus by Levy's characterization of Brownian motion, you may come up with some $\sigma W_t$, where $\sigma$ is a fixed constant and $W_t$ is another Wiener process, such that
$$
\sigma W_t=\sigma_1W_t^1+\sigma_2W_t^2
$$
almost surely in the sense of their trajectories. Here "in the sense of their trajectories" means that, intuitively, the trajectories of $\sigma W_t$ and of $\sigma_1W_t^1+\sigma_2W_t^2$ are absolutely indistinguishable.
Interestingly, if this is the case, it would remain somewhat wired as $\sigma_1W_t^1+\sigma_2W_t^2$ appears in the SDE. After all, that Brownian motions are included in an equation is not because we want to include them, either for fun or for else reasons. Instead, it is that we hope these Brownian motions have practical meanings, either financially or physically. If one Brownian motion $\sigma W_t$ suffices, there is intuitively unnecessary to bring in two, just like $\sigma_1W_t^1+\sigma_2W_t^2$, because practically you can owe multiple stochastic factors to a single factor, unless it is a must to tell these stochastic factors apart.
By contrast, if you have more than one SDE, distinguishable stochastic processes would become meaningful. For example, consider
\begin{align}
{\rm d}S_t^1&=\mu_t^1{\rm d}t+\sigma_t^{11}{\rm d}W_t^1+\sigma_t^{12}{\rm d}W_t^2,\\
{\rm d}S_t^2&=\mu_t^2{\rm d}t+\sigma_t^{21}{\rm d}W_t^1+\sigma_t^{22}{\rm d}W_t^2,
\end{align}
which can be reformatted as
$$
{\rm d}\left(
\begin{array}{c}
S^1\\
S^2
\end{array}
\right)_t=\left(
\begin{array}{c}
\mu^1\\
\mu^2
\end{array}
\right)_t{\rm d}t+\left(
\begin{array}{cc}
\sigma^{11}&\sigma^{12}\\
\sigma^{21}&\sigma^{22}
\end{array}
\right)_t{\rm d}\left(
\begin{array}{c}
W^1\\
W^2
\end{array}
\right)_t.
$$
You may see that $S_t^1$ and $S_t^2$ are now "entangled". You may use the same trick as above to deal with each SDE separately. But if you choose to do so with, say, $S_t^1$, you will then have no idea about $S_t^2$, because the latter depends on $\sigma_t^{21}{\rm d}W_t^1+\sigma_t^{22}{\rm d}W_t^2$, rather than some constructed $\sigma_t{\rm d}W_t$ (think about this: if you let $\sigma_t{\rm d}W_t=\sigma_t^{11}{\rm d}W_t^1+\sigma_t^{12}{\rm d}W_t^2$, then how can you recover $\sigma_t^{21}{\rm d}W_t^1+\sigma_t^{22}{\rm d}W_t^2$ by using $\sigma_t{\rm d}W_t$ solely?).
To sum up, if you have only one SDE, your method works, and works well. In this case, however, the SDE itself witnesses further simplification. By contrast, if you have more than one SDE, your method still works for each SDE, but it contributes little to the whole SDE system.
To see that $W_{2,t}$ is a martingale wrt $\mathcal{F}^{W_2}_s:=\sigma(\{W_{2,h},h\leq s\})$ for $s<t$ we see that, since $W_{2,t}-W_{2,s}\sim\mathcal{N}(0,t-s)$ is independent from $\mathcal{F}^{W_2}_s$ and $W_{2,s}$ is $\mathcal{F}^{W_2}_s$-measurable so it can be 'taken out' from the conditional expectation we get
$$E[W_{2,t}|\mathcal{F}^{W_2}_s]=E[(W_{2,t}-W_{2,s})+W_{2,s}|\mathcal{F}^{W_2}_s]=W_{2,s}E[1|\mathcal{F}^{W_2}_s]=W_{2,s}$$
To see that $W_{2,t}$ is not a $\mathcal{F}_t^{W_1}$-martingale, since $W_{2,s}$ is independent from $W_{1,s}$ we get
$$E[W_{2,t}|\mathcal{F}_s^{W_1}]=E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$$
which is not a.s. equal to $W_{2,s}$. Indeed for $A \in \mathcal{F}_s^{W_1}$
$$\begin{aligned}E[E[W_{2,s}|\mathcal{F}_s^{W_1}]\mathbb{I}_A]&=E[W_{2,s}\mathbb{I}_A]=\\&=E[W_{2,s}]E[\mathbb{I}_A]=\\
&=E[E[W_{2,s}]\mathbb{I}_A]\end{aligned}$$
so $E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$ a.s.
Best Answer
Basically, you can simply apply Itô's formula: $$\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^2 \\ &\quad +\frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad + \int_0^t f_{yy}(W_s^1,W_s^2) \, \underbrace{dW_s^2 dW_s^2}_{ds} \bigg). \end{align*}$$
If we choose $f(x,y) := \exp(\sigma(x+y))$ and take the expectation on both sides, it follows that
$$\mathbb{E}e^{\sigma (W_t^1+W_t^2)}-1 = \sigma^2(\varrho+1) \int_0^t \mathbb{E}e^{\sigma (W_s^1+W_s^2)} \, ds.$$
Setting $\varphi(t) := \mathbb{E}e^{\sigma (W_t^1+W_t^2)}$ we find that $\varphi$ is a solution to the ordinary differential equation
$$\varphi'(t) = \sigma^2 (\varrho+1) \varphi(t) \qquad \varphi(0)=1.$$
This ODE can be solved explicitly,
$$\varphi(t) = \exp(\sigma^2(\varrho+1)t).$$
Combining the results, we conclude
$$\mathbb{E}(X_t\cdot Y_t) = X_0 Y_0 e^{2\mu t} e^{\sigma^2 \varrho t}.$$