To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation.
Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form
$$dX_t = b(t) \, dt + \eta(t) \, dW_t$$
where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an $m$-dimensional Brownian motion. Then Itô's formula states that
$$\begin{align*} f(X_t)-f(X_0) &= \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s \end{align*}$$
for $f \in C^2$ where
$$\begin{align*} \int_0^t f'(X_s) \, dX_s &:= \int_0^t f'(X_s) \eta_s \, dW_s + \int_0^t f'(X_s) b(s) \, ds \\ d\langle X \rangle_s &:= \sum_{j=1}^m \eta_j^2(s) \, ds. \end{align*}$$
In your setting, we have $$b(t) = \left(\mu_t - \frac{1}{2} \sum_{j=1}^m \sigma_t^j \right)^2 \qquad \eta(t) := (\sigma_t^1,\ldots,\sigma_t^m).$$
Applying Itô's formula for $f(x) := \exp(x)$ shows that $(S_t)_{t \geq 0}$ is a solution to the given SDE.
Writing
\begin{align}
E_t&:=\mathcal{E}(X)_t\,,\\
F_t&:=y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s
\end{align}
Applying Ito to $Z_t=E_tF_t$ gives
\begin{align}
dZ_t&=E_t\,dF_t+F_t\,dE_t+d\langle E,F\rangle_t\\[3mm]
&=dY_t-d\langle X,Y\rangle_t+F_t\,E_t\,dX_t+d\langle X,Y\rangle_t\\[3mm]
&=dY_t+Z_\,dX_t\,
\end{align}
as expected. Here I used
\begin{align}
dF_t&=E_t^{-1}\,dY_t-E_t^{-1}\,d\langle X,Y\rangle_t\,,\\[3mm]
dE_t&=E_t\,dX_t\,,\\[3mm]
d\langle E,F\rangle_t&=E_t E_t^{-1}\,d\langle X,Y\rangle_t=d\langle X,Y\rangle_t\,,\\[3mm]
d\langle X,\langle X,Y\rangle\rangle_t&=0\,.
\end{align}
Regarding uniqueness:
Let $Z$ and $Z'$ be solutions with $Z_0=Z'_0$. Then $\widetilde{Z}:=Z-Z'$ is a solution of
$$
d\widetilde{Z}_t=\widetilde{Z}_t\,dX_t
$$
with $\widetilde{Z}_0=0\,.$ Since this equation has a unique strong solution
of the form
$$
\widetilde{Z}_t=\widetilde{Z}_0\,e^{X_t-\frac{1}{2}\langle X\rangle_t}
$$
it follows that $\widetilde{Z}_t\equiv 0$.
Best Answer
So from Itô (on $S_t$ and $ln$ to get $Y_t$'s SDE), you get the following : $$Y_t=Y_0 -\frac{1}{2} \int_0^t Z_s^2 ds + \int_0^t Z_sdB_s$$ with $Z_s= \frac{\eta_s}{e^{Y_s}}$.
Now observe the following : $$exp (Y_T)=exp (X)$$ so $$X=Y_T=Y_0 -\frac{1}{2} \int_0^T Z_s^2 ds + \int_0^T Z_sdB_s$$ (and by the way you can see that taking $t=0$, that $Y_0= ln (\mathbb E[exp(X)])$ but we won't need that).
We are almost done now take $Y_t +(X-X)$ and the following comes : $$Y_t=Y_t +(X-X)=X - (\frac{1}{2}\int_0^t Z_s^2 ds - \frac{1}{2}\int_0^T Z_s^2 ds) +(\int_0^t Z_sdB_s-\int_0^T Z_sdB_s)$$ or $$Y_t= X + \frac{1}{2}\int_0^T Z_s^2 ds -\int_t^T Z_sdB_s$$
So I think there is typo in your exercise as I have found a minus term on the anticipative term of Brownian integral in the final representation of $Y$ unless mistaken.