Consider $M= \left(M\right )_{t \geq0}, \ N=\left(N\right) _{t \geq0} \in \mathcal M_{c,loc} $ starting both from zero, such that, a.e.$ \langle M \rangle_t =\langle N \rangle_t, \ \forall t\geq 0$.
Have $M$ and $N$ the same law? If not give a contre exemple.
No. $M$ and $N$ need not have the same distribution. For example, take $$ Z_t = \exp\left\{B_t - \frac{t}{2}\right\},\qquad M_t = Z_t - 1,\qquad N_t = 1 -Z_t. $$
The continuous martingales $M$ and $N$ have the same quadratic variation $\langle Z\rangle_t$, but since $Z_t$ tends to $0$ as $t \to +\infty$ a.s. they cannot have the same distribution.
Proof that $Z_t \xrightarrow[t\to\infty]{} 0$ a.s.
We know that $\dfrac{B_t}{t} \xrightarrow[t\to\infty]{} 0$ as a consequence of the strong law of large numbers (or the law of the iterated logarithm). From this we deduce that $$Z_t = \exp\left\{t \left(\frac{B_t}{t}-\frac{1}{2}\right) \right\}\xrightarrow[t\to\infty]{} 0\quad\text{a.s.} $$