Ito formula for jump-Diffusion

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 $$ f(S_t) - f(S_{t-}) = f(j_t S_{t-}) - f(S_{t-}) = \ln(j_t) + \ln(S_{t-}) - \ln(S_{t-}) $$

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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, $$ a_t \frac{\partial f(S_t, t)}{\partial x} dt = \mu S_t \frac{1}{S_t} dt = \mu dt $$