1. We need the set of all possible values of $f(x)=\sin x-\sqrt{3}\cos x+1.$ Since $f:R\to S$ is onto, the codomain $S$ must equal the range of $f$.

2. First find the range of $\sin x-\sqrt{3}\cos x.$ Write it in the form $R\sin(x-\phi)$.

Let $\sin x-\sqrt{3}\cos x = R\sin(x-\phi)=R(\sin x\cos\phi-\cos x\sin\phi).$ Comparing coefficients: $R\cos\phi=1, \qquad R\sin\phi=\sqrt{3}.$ So, $R=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2.$ Hence, $\sin x-\sqrt{3}\cos x = 2\sin(x-\phi)$ for some $\phi$, and therefore its range is $[-2,2].$

3. Now add $1$ to this expression: $f(x)=\left(\sin x-\sqrt{3}\cos x\right)+1.$ So the range shifts by $+1$: $[-2,2]+1=[-1,3].$

4. Since $f:R\to S$ is onto, we must have $S=[-1,3].$

5. Check options:

• A: $[-1,3]$ ✅
• B: $[-1,1]$ ❌
• C: $[0,1]$ ❌
• D: $[0,3]$ ❌

Therefore, the correct answer is A.