In a book is a exercise to prove Yor's formula for stochastic exponential, i.e.
$\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}=\mathcal{E}(X)\mathcal{E}(Y)$
where $\mathcal{E}(X)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}$. Now if we would define the stochastic exponential as $\exp{(X_t-\frac{1}{2}\langle X\rangle)}$, then the above is simple algebra:
$\mathcal{E}(X)_t\mathcal{E}(Y)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}=\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}$
Using that $\langle X,Y\rangle=\frac{1}{2}(\langle X+Y\rangle-\langle X\rangle -\langle Y\rangle$ we would get:
$\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}=\exp{(X_t+Y_t+\langle X,Y\rangle -\frac{1}{2}\langle X+Y\rangle)}=\exp{(X_t+Y_t-\frac{1}{2}\langle X+Y\rangle)}\exp{(\langle X,Y\rangle)}=\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}$
If I would define the stochastic exponential as the (unique) solution $Z_t$ of the SDE $Z_tdX_t=dZ_t$. Then I guess I have to use Itô, to prove the statement, but how exactly? However, if we define the stochastic exponential as above, then my conclusion would be correct?