Suppose that $S$, the relevant boundary component of $M$, is a torus.  Suppose that $G$ is the given essential two-sphere in the filled manifold $N$.  We isotope $G$ to have minimal intersection with $S$.  All intersections $S \cap G$ are essential simple closed curves in $S$.  So all of them are parallel.  Set $F = M \cap G$.  As you argue, $F$ is incompressible.  

Suppose, for a contradiction, that $F$ is boundary compressible.  Let $B \subset M$ be the given bigon boundary compressing $F$.  Let $\beta = B \cap F$ and let $\beta' = B \cap S = B \cap \partial M$.  

Let $\alpha$ and $\alpha'$ be the curves of $S \cap G$ which meet the *corners* of $B$ - that is, the points $\beta \cap \beta'$. If $\alpha = \alpha'$ then $\beta'$ is an inessential arc in $(S, \alpha)$.  So $\beta'$ cuts a bigon out of $S - G$.  Thus $B \cup B'$ is a compression of $F$, a contradiction. 

Suppose instead that $\alpha$ and $\alpha'$ are distinct.  Thus they co-bound an annulus $A$ in $S$.  We choose $A$ so that $A \cap G = \alpha \cap \alpha'$. That is, there are no more curves of $G$ in $A$.  Let $D$ be the disk obtained by boundary compressing $A$, along $B$, into $M$. This disk $D$ is a compressing disk for $F$ (or we can further reduce $G \cap S$).  This is the desired contradiction.

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If $S$ is not a torus, the argument is more involved.  I will give details in my answer to your previous [question][1].


  [1]: https://mathoverflow.net/questions/476127/is-it-possible-to-fill-a-boundary-component-of-an-irreducible-3-manifold-using-a/476128#476128