My attempt at a solution is:
For $s<t$
$E[X_t|\mathcal{F}_s]=X_s \\ E[e^{bt}\cdot e^{\sigma}w_t]=e^{bs}e^{\sigma w_s}$
$e^{b(t-s)}E[e^{\sigma (w_t-w_s+w_s)}]=e^{\sigma w_s}$
$e^{b(t-s)}\cdot e^{\sigma w_s} E[e^{\sigma (w_t-w_s)}] = e^{\sigma w_s}$
The expected value is the Moment Generating Function for the variable $(w_t-w_s)$, and we simplify the term $e^{\sigma w_s}$ at each side
$e^{b(t-s)}\cdot e^{\frac{\sigma ^2}{2} (w_t-w_s)}=1$
Therefore
$b = -\frac{\sigma ^2 (w_t-w_s)}{2(t-s)}$
I'm pretty sure this is wrong, since according to my answer $b$ would depend from $(w_t-w_s)$. I was wondering if anyone has any pointers or other methods I can try.
As saz pointed out, the application of the MGF is not correct. I'm not sure if that made the solution clear staright away, but here's my go at a more rigorous answer. We have as given, adding implied initial value of $X_0=1$ $$X_t=X_0e^{bt+\sigma W_t}$$ Let some variable $Y$ be the exponent, that is $Y=bt+\sigma W_t$. The moment generating function of a normal r.v. $Y\sim N(b,\sigma^2)$ is $$M_Y(s)=E(e^{sY})=e^{bs+\frac{\sigma^2 s^2}{2}},\quad -\infty<s<\infty$$ So, for a BM with drift, as here, we have $Y(t)\sim N(bt,\sigma^2t)$, leading to $$M_{Y(t)}(s)=E(e^{sY(t)})=e^{bts+\frac{\sigma^2ts^2}{2}},\quad -\infty<s<\infty$$ So, using the MGF, the first moment is given by $$E(X(t))=E(X_0e^{Y(t)})=X_0M_{Y(t)}(1)=X_0e^{bt\cdot1+\frac{\sigma^2t\cdot1^2}{2}}<=>$$ $$E(X(t))=X_0e^{(b+\frac{\sigma^2}{2})t}$$ Thus, to satisfy the martingale condition, we see that we must have $$b=-\frac{\sigma^2}{2}$$