Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. Hence, (needless to say) $X$ and $Y$ are both semimartingales. Since $(x,y) \mapsto e^{x+y}$ is twice continuously differentiable in all its arguments, Ito's formula applies, i.e.
$$dZ_t = Z_tdX_t + Z_tdY_t + \frac{1}{2}Z_td[X,X]_t+ \frac{1}{2}Z_td[Y,Y]_t + Z_td[X,Y]_t$$
You can check the following identities yourself.
$$dX_t = -\beta_tdW_t$$
$$dY_t = -\frac{1}{2}\beta_t^2dt$$
$$d[X,X]_t = \beta^2_tdt$$
$$d[X,Y]_t = d[Y,Y]_t = 0$$
Substituting these identities into the SDE above,
$$dZ_t = -Z_t\beta_tdW_t - Z_t\frac{1}{2}\beta_t^2dt + Z_t\frac{1}{2}\beta^2_tdt$$
Hence, $dZ_t = -Z_t\beta_tdW_t$. I mentioned above that there must be a condition on $\beta$ that makes $X$ a local martingale. The most general condition for this is that
$$\int_0^t\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\triangle)$$
for every $t < \infty$. We will require the same for $Z$ to be a local martingale, i.e.
$$\int_0^tZ_s^2\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\square)$$
for every $t < \infty$. Now fix an arbitrary $0 < t < \infty$. Let $\Omega_t$ denote the full measure set where $(\triangle)$ is true. Let $\Omega_c$ be the full measure set where $Z$ is continuous. Fix $\omega \in \Omega_c \cap \Omega_t$. Since $s \mapsto Z_s(\omega)$ is continuous, $s \mapsto 1_{[0,t]}(s)Z_s(\omega)$ is bounded. Then,
$$\int_0^tZ_s(\omega)^2\beta_s(\omega)^2\,ds \leq \sup_{s \in [0,t]}Z_s(\omega)^2 \int_0^t\beta_s(\omega)^2\,ds < \infty$$
Since $\Omega_c \cap \Omega_t$ has full measure, $(\square)$ is true. So then, $Z$ is a local martingale.
Itô's formula shows that
$$M_t = 1+ \int_0^t f(s) M_s \, dW_s \tag{1}$$
and this implies, in particular, that $(M_t)_{t \geq 0}$ is a local martingale. (Note that $(M_t)_{t \geq 0}$ has continuous sample paths, and therefore the stochastic integral on the right-hand side is well-defined.) On the other hand, it follows from the very definition that $M_t \geq 0$ for each $t \geq 0$. Since any non-negative local martingale is a supermartingale (see e.g. this question for details), we conclude that $(M_t)_{t \geq 0}$ is a supermartingale. Thus,
$$\mathbb{E}(M_t) \leq \mathbb{E}(M_0) \leq 1$$
which implies
$$\mathbb{E} \exp \left( \int_0^t f(s) \, dW_s \right) \leq \exp \left( \frac{1}{2} \int_0^t f(s)^2 \, ds \right)$$
for each $t \geq 0$. Replacing $f$ by $2f$ we find in particular that
$$\mathbb{E} \left| \exp \left( \int_0^t f(s) \, dW_s \right) \right|^2 = \mathbb{E}\exp \left( \int_0^t 2f(s) \, dW_s \right) \leq \exp \left( 2 \int_0^t f(s)^2 \, ds \right) < \infty,$$
and so
$$\mathbb{E}(M_t^2) \leq \exp \left( \int_0^t f(s)^2 \, ds \right).$$
Using this estimate and the fact that $f$ is deterministic, we can easily check that the stochastic integral $\int_0^t f(s) M(s) \, dW_s$ is a true martingale, and now $(1)$ shows that $(M_t)_{t \geq 0}$ is a martingale.
Best Answer
This may not be exactly what you're looking for, but here's a proof which at least uses the specific form of the process (being the exponential local martingale of an integral of a square-integrable deterministic process with respect to a Brownian motion):
First off, in order to make things fit better with the standard framework, I'll assume that $\phi$ is a mapping from $[0,\infty)$ to $\mathbb{R}$ with the property that $\int_0^t \phi(s)^2ds$ is finite for all $t\ge0$. Define $N_t = \int_0^t \phi(s)dW_s$, we then have $M_t = \mathcal{E}(N)_t$, the exponential local martingale, and the objective is to prove that $\mathcal{E}(N)$ is a martingale (instead of just a local martingale). By some classical results, $\mathcal{E}(N)$ is a nonnegative supermartingale, and it is a martingale if and only if $E\mathcal{E}(N)_t = 1$ for all $t\ge0$. I don't know where you actually can find a proof of these claims, but they follow from applications of the optional sampling theorem and Fatou's lemma. The conclusion is that we need to show $E\mathcal{E}(N)_t=1$ for all $t\ge0$.
To do so, first note that $[N]_t = \int_0^t \phi(s)^2 ds$. We apply Itô's formula:
$\mathcal{E}(N)^2_t = 1 + 2\int_0^t \mathcal{E}(N)_sd\mathcal{E}(N)_s+[\mathcal{E}(N)]_t\\ =1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_sd[N]_s\\ =1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_s\phi(s)^2ds.$
As $\mathcal{E}(N)$ and $\mathcal{E}(N)\cdot \mathcal{E}(N)$ are continuous local martingales, they are locally bounded. Let $(T_n)$ be a localising sequence of stopping times such that both $\mathcal{E}(N)^{T_n}$ and $(\mathcal{E}(N)\cdot \mathcal{E}(N))^{T_n}$ are bounded martingales. By the martingale property of $(\mathcal{E}(N)\cdot\mathcal{E}(N))^{T_n}$, we then obtain
$E\mathcal{E}(N)_{t\land T_n}^2 = 1 + E\int_0^{t\land T_n}\mathcal{E}(N)_sd\mathcal{E}(N)_s + E\int_0^{t\land T_n}\mathcal{E}(N)^2_s\phi(s)^2 ds\\ =1 + E \int_0^t \mathcal{E}(N)^2_s\phi(s)^21_{[0,T_n]}(s)ds \le 1 + \int_0^t E\mathcal{E}(N)^2_{s\land T_n}\phi(s)^2ds$
where we also applied Tonelli's theorem and nonnegativity of $\mathcal{E}(N)$. Now consider a fixed $n\ge1$ and define $g_n(t) = E\mathcal{E}(N)^2_{t\land T_n}$. The above then states that
$g_n(t) \le 1+ \int_0^t g_n(s)\phi(s)^2ds$.
By a classical analysis lemma, Gronwall's lemma (See the Wikipedia article on Gronwall's inequality), we then obtain $g_n(t) \le \exp(\int_0^t\phi(s)^2ds)$. In other words, we have now shown
$E\mathcal{E}(N)^2_{t\land T_n}\le \exp(\int_0^t \phi(s)^2ds)$
for all $n\ge1$ and $t\ge0$. Now fix $t\ge0$. By the above, the family $(\mathcal{E}(N)_{t\land T_n})_{n\ge1}$ is then bounded in $\mathcal{L}^2$, therefore uniformly integrable. Furthermore, $\mathcal{E}(N)_{t\land T_n}$ converges almost surely to $\mathcal{E}(N)_t$. Combining this with uniform integrability, $\mathcal{E}(N)_{t\land T_n}$ converges in $\mathcal{L}^1$ to $\mathcal{E}(N)_t$, and so the means also converge. And as $\mathcal{E}(N)^{T_n}$ is a martingale, $\mathcal{E}(N)_{t\land T_n}=1$. We conclude that $E\mathcal{E}(N)_t = 1$, and so $\mathcal{E}(N)$ is a martingale.