You can look at your process $X_{t}$ as a two dimensional stochastic process
$$Y_{t}=\left[\begin{array}{cc}X_{t}\\ \eta_{t}\end{array}\right]$$
Then
$$dY_{t}=\left[\begin{array}{cc}dX_{t}\\ d\eta_{t}\end{array}\right]=\left[\begin{array}{cc}b(X_t)+\lambda\eta_{t}\sigma(X_{t})\\ \lambda\eta_{t}\end{array}\right]dt+\left[\begin{array}{cc}\alpha\sigma(X_t)&0\\ 0&\alpha\end{array}\right]\left[\begin{array}{cc}dW_{t}\\ dW_{t}\end{array}\right]$$
and the infinitesimal generator is of the form
$$LV(y)=LV(x,\eta)=\left(b(x)+\lambda\eta \sigma(x)\right)V'_{x}(x,\eta)+\lambda\eta V'_{\eta}(\eta,x)$$
$$+\frac{1}{2}\alpha^{2}\sigma^{2}(x)V''_{xx}(x,\eta)+\alpha^{2}\sigma(x)V''_{x\eta}(x,\eta)+\frac{1}{2}\alpha^{2}V''_{\eta\eta}(x,\eta)$$
By the way, the infinitesimal generator of an Ornstein-Uhlenbeck process of the form
$$d\eta_{t} = \lambda\eta_{t} dt + \alpha dW_{t}$$
is
$$LV(\eta)=\lambda \eta V'(\eta) + \frac{\alpha^2}{2}V''(\eta)$$
The problem here is that you are using the Ito formula for Lévy processes – your process is not a Lévy process, but a semimartingale that is driven by a Lévy process (read the 8.3.4 – "Ito Formula for Seminartingales" in the Tankov book). For example, at a point $t$ where there is a jump $j_t$, your process doesn't satisfy
$$
S_t = S_{t-} + j_t
$$
but rather
$$
S_t = j_tS_{t-}
$$
More precisely, you process is a Dolean-Dades exponential of a Lévy process.
So you have to use Ito for semimartingales (proposition 8.19 in Tankov), i.e.
$$
\begin{align}
f(t, X_t) - f(0, X_0) &= \int^t_0 \frac{\partial f}{\partial s}(s, X_s) ds + \int_0^t\frac{\partial f}{\partial x}(s, X_s) dX_s \\
&+ \frac{1}{2} \int^t_0 \frac{\partial^2 f}{\partial x^2}(s, X_s) d[X, X]_s^c \\
&+\sum_{s \leq t, \Delta X_s \neq 0} [f(s, X_s) - f(s, X_{s-}) - \Delta X_s\frac{\partial f}{\partial x}(s, X_s)]
\end{align}
$$
where $[X, X]_s^c$ is the quadratic variation of the continuous part of $X$.
This formula is less straight-forward to apply, because for example if you insert $S_t$ as $X_t$ and use $\log$ as $f$, then the second term on RHS is
$$
\int_0^t\frac{\partial f}{\partial x}(s, S_s) dS_s = \int_0^t\frac{1}{S_t}S_t(\mu ds + \sigma dW_s + dj) = \mu t + \sigma W_t + \int^t_0dj
$$
How do you interpret the last integral in this situation? It is actually an integral against the jump measure of $S_t$, $J_S(x, t)$. This is pure hand-waving, but for now I just argue that this is the sum of jumps up to time $t$:
$$
\int^t_0dj = \int_0^t \int_\mathbb R xJ_S(ds, dx) = \sum_{s\leq t, \Delta S_s \neq 0} j_s
$$
This part will cancel against the term $- \Delta X_s\frac{\partial f}{\partial x}(s, X_s)$ inside the sum in the Ito for semimartingales so that essentially, you will be left with the usual Ito for Lévy processes, except that the jump part will look like
$$
f(X_t) - f(X_{t-})
$$
instead of
$$f(X_{t-} + \Delta X_t) - f(X_{t-})$$
Thus, at a point where there is a jump, you get
$$
f(S_t) - f(S_{t-}) = f(j_t S_{t-}) - f(S_{t-}) = \ln(j_t) + \ln(S_{t-}) - \ln(S_{t-})
$$
As for your solution attempt I'm not sure what you tried to do there – it seems you are trying to derive an Ito formula of your own. I suggest using the original formula with the said modification for the jumps.
Your process corresponds to
$$
S_t = S_0 + \int^t_0 S_s \mu ds + \int^t_0 \sigma S_s dW_s + \sum S_s j_s
$$
from which you can read off $a_t$, $b_t$, and then plug into the Ito formula. For example,
$$
a_t \frac{\partial f(S_t, t)}{\partial x} dt = \mu S_t \frac{1}{S_t} dt = \mu dt
$$
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