Let $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space and $(\mathcal F_t)_{t\ge0}$ be a complete filtration on $(\Omega,\mathcal A,\operatorname P)$.
Let $(M_t)_{t\ge0}$ be a local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ and $$X^\sigma:=\sigma M-\frac{\sigma^2}2[M]\;\;\;\text{for }\sigma\in\mathbb R.$$ By the Itō formula, $$N^\sigma:=e^{X^\sigma}=N_0+\sigma N\cdot M\tag1$$ is a local $\mathcal F$-martingale for all $\sigma\in\mathbb R$.
Assume $M$ is right-continuous. I want to show that $$\operatorname P\left[\sup_{s\in[0,\:t]}X^\alpha_s\ge\alpha\beta\right]\le e^{-\alpha\beta}\;\;\;\text{for all }t\ge0\text{ and }\alpha,\beta>0.\tag2$$
If $M$ is a (strict) $\mathcal F$-martingale, then we can show that $N^\sigma$ is an $\mathcal F$-martingale as well and hence we obtain $$\operatorname P\left[\sup_{s\in[0,\:t]}\ge a,[M]_t\le b\right]\le\operatorname P\left[\sup_{s\in[0,\:t]}N^\sigma_s\ge e^{\sigma a-\frac{\sigma^2}2b}\right]\le e^{\frac{\sigma^2}2b-\sigma a}\tag3$$ by Doob's martingale inequality for all $t\ge0$ and $a,b>0$.
Are we able to show $(3)$ even when $M$ is only a local $\mathcal F$-martingale? If that would be the case, we might be able to show $(2)$ by noting that $$M_s-\frac\sigma2[M]_s\ge a-\frac{\sigma^2}b\Leftrightarrow N^\sigma_s\ge e^{\sigma a-\frac{\sigma^2}2b}\tag4$$ for all $s\ge0$ and $a,b,\sigma\in\mathbb R$.
Assuming $M$ is continuous, the process $Y:=\exp(X^\alpha)$ is a non-negative local martingale, with initial value $1$; in particular, $Y$ is a non-negative supermartingale. You then have, taking $T$ to be the first time $X^\alpha$ is at least $\alpha\beta$: $$ 1=\Bbb E[Y_0]\ge \Bbb E[Y_T; T<\infty]\ge \exp(\alpha\beta)\cdot\Bbb P[T<\infty]. $$Finally, notice that $\{T<\infty\}=\{\sup_{s\ge 0}X_s^\alpha\ge\alpha\beta\}\supset \{\sup_{0\le s\le t}X^\alpha_s\ge\alpha\beta\}$.