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.
Best Answer
According to a result of R.M. Dudley [Wiener functionals as Itô integrals, Ann. Probability 5 (1977), no. 1, 140–141], given an $\mathcal F_1$-measurable random variable $X$, there is a predictable $(H_t)$ with $\int_0^t H_s^2 ds<\infty$ such that $X=\int_0^1 H_s dB_s$, a.s. Adapting this in the obvious way, one has predictable processes $(H_t)$ and $(K_t)$ such that (i) $H_t=0$ for $1/2<t\le 1$, (ii) $K_t=0$ for $0\le t\le 1/2$, and (iii) $\int_0^1 H_s dB_s =1 =\int_0^1 K_s dB_s$, a.s. The difference $\sigma_t:=H_t-K_t$ is then non-trivial but $\int_0^1 \sigma_s dB_s =0$, a.s. Take $M_0=0$ and you have a counterexample: $M_t=\int_0^t \sigma_s dB_s$ is a continuous local martingale, but not a martingale.