Consider the regular function given by $S_1/S_0$ on the projective sphere $\mathbb{P}^2$ over a field $k$ (We can assume algebraically closed, if it's needed for some reason). I'm just worried, is the domain of definition just $S_0\neq 0$? Or does the domain extend a little further?
The reason I'm uncertain is because I know the function $(1-y)/x$ on the circle $x^2+y^2=1$ extends beyond just $x\neq 0$, by the manipulation $$ \frac{1-y}{x}=\frac{(1-y)(1+y)}{x(1+y)}=\frac{1-y^2}{x(1+y)}=\frac{x^2}{x(1+y)}=\frac{x}{1+y} $$ to include the point $(0,1)$ as well. Am I overlooking anything like that here? Thank you.
If you mean $\mathbb{P}^1_k$ as in the comments, there's not much to show as $\{S_0=0\}$ is just one point.
If you mean $\mathbb{P}^2_k$ and $\frac{S_1}{S_0}$ is regular at $[x_0: x_1: x_2]$ with $S_0(x_0)= 0$, we need $S_1(x_1)=0$ as well. (For example, this happens in the example you gave above.)
This means, the only point we would need to be worried about is $[0:0:1]$. However, $\frac{S_1}{S_0}$ does not extend to this point. Intuitively, the zeroes of $S_0$ extend in a different direction as the zeros of $S_1$, which can't happen in the case of a curve, as in your example above.
To prove this, we can for example, take an affine chart to reduce the problem to showing that $\frac{X_1}{X_0}\neq k[X_0,X_1]_{\langle X_0,X_1\rangle}$. One way to go is to note that, if $\frac{X_1}{X_0}$ is in $k[X_0,X_1]_{\langle X_0,X_1\rangle}$, then $\langle X_0,X_1\rangle$ is generated by 1 element, so the dimension of $k[X_0,X_1]$ is at most 1, which is a contradiction.