Let the price $S_t$ of an asset satisfy $dS_t = \alpha (\mu - \ln{S_t})S_tdt + \sigma S_t dW_t$, where $W_t$ is a Brownian motion.
I managed to show that $x_T = x_te^{-b(T-t)} + \frac{a}{b}(1 - e^{-b(T-t)}) + \sigma e^{bT} \int_{t} ^{T} e^{bs}dW_s$ for a later time $T > t$, where $x_t = \ln S_t$, $a= \alpha \hat{\mu}$ for a choice of $\hat{\mu}$ and $b = \alpha$.
Now I need to use this in order to find $\mathbb{E}_{t}[S_T]$. I don't really know how to proceed since I have the equation in terms of the logarithm of $S_T$...
As I guess that you algready found out, Itô's lemma for $X_t := \ln S_T$ leads to an Ornstein-Uhlenbeck process, which is given by $$ X_t = X_{t_0} + \left(1-\frac{\sigma^2}{2\alpha\mu}\right)\left(1-e^{-\alpha\mu(t-t_0)}\right) + \sigma\int_{t_0}^t e^{-\alpha\mu(t-s)}\,\mathrm{d}W_s. $$ Since $$ \mathrm{Var}\left[\int_{t_0}^t e^{-\alpha\mu(t-s)}\,\mathrm{d}W_s\right] = \int_{t_0}^t e^{-2\alpha\mu(t-s)}\,\mathrm{d}s = \frac{1-e^{-2\alpha\mu(t-t_0)}}{2\alpha\mu} $$ by Itô's isometry, we deduce that $X_t$ (conditioned on $X_{t_0}$) follows a normal law, namely $$ X_t-X_{t_0} \sim \mathcal{N}\left(\left(1-\frac{\sigma^2}{2\alpha\mu}\right)\left(1-e^{-\alpha\mu(t-t_0)}\right), \frac{\sigma^2}{2\alpha\mu}\left(1-e^{-2\alpha\mu(t-t_0)}\right)\right). $$ In consequence, $S_t = S_{t_0}e^{X_t-X_{t_0}}$, with $S_{t_0} = e^{X_{t_0}}$, is a log-normal variable, hence $$ \mathbb{E}[S_t|S_{t_0}] = S_{t_0}\mathbb{E}[e^{X_t-X_{t_0}}] = S_{t_0}e^{\mathbb{E}[X_t-X_{t_0}] + \frac{1}{2}\mathrm{Var}[X_t-X_{t_0}]} = \ldots $$