Converting a regular expression to a context-free grammar
This CFG is indeed correct, but can be greatly simplified.

First, there is no need to create a new variable that would derive only a terminal value, that is - $s_1,s_2,s_5,s_6$ are not required.
For a similar reason, we don't need $s_9$.
Another simplification is that $(a\cup b)^*$ was already generated by $s_4$. There is no need to create $s_7$ and $s_8$ for this usage

Considering those things, the CFG will be reduced to:

$S\rightarrow A\space|\space abA$
$A \rightarrow AB\space|\space\epsilon$
$B \rightarrow a\space |\space b$

Notice also that $L(ab(a\cup b)^*)\subseteq L((a\cup b)^*)$, and thus:
$L((a\cup b)^*\cup ab(a\cup b)^*) = L((a\cup b)^*) \cup L(ab(a\cup b)^*) = L((a\cup b)^*)$
And thus an even smaller CFG will be:

$S \rightarrow A$
$A \rightarrow AB \space | \space \epsilon$
$B \rightarrow a\space | \space b$

Essentially we removed one unnecessary derivation rule. This can be simplified a bit further, go ahead and give it a try :)