Analytic continuation of a holomorphic function $f$ is the same along any closed curve in $D$ implies $f$ extends to $D$?

Question

Let $D\subset\mathbb{C}$ be a domain and let $f$ be the germ of a holomorphic function which admits unrestricted analytic continuation in some $E$ that contains $D$. Furthermore, suppose that analytic continuation of $f$ along any closed curve $\gamma:[0,1]\to D$ yields $f(\gamma(0))=f(\gamma(1))$, i.e. the value at the end of the curve is the same as the value at the starting point of the curve. Does this imply that $f$ can be analytically continued to all of $D$?

I had the following situation in mind: $\sqrt{1+z}:=e^{\tfrac{1}{2}\log(1+z^2)}$ admits unrestricted analytic continuation in $\mathbb{C}\setminus\{-i,i\}$, and furthermore for any closed loop $\gamma:[0,1]\to\mathbb{C}\setminus\{ix\in i\mathbb{R}\,:\,|x|<1\}$, we have that $$ \frac{1}{2\pi i}\int_{1+\gamma^2}\frac{dw}{w}=\frac{1}{2\pi i}\int_{\gamma}\frac{2z}{z^2+1}dz=\text{Ind}(\gamma,i)+\text{Ind}(\gamma, -i). $$ (Here $\{ix\in i\mathbb{R}\,:\,|x|<1\}$ is the "slit" in the plane connecting $i$ and $-i$.) But since $i$ and $-i$ belong to the same connected component of $\{ix\in i\mathbb{R}\,:\,|x|<1\}$, we see that $\text{Ind}(\gamma,i)=\text{Ind}(\gamma,-i)=k$ for some nonnegative integer $k$ and hence the curve $1+\gamma^2(t)$ has even winding number about the origin. Therefore $$ \log(1+\gamma(1)^2)=\log(1+\gamma(0)^2)+4\pi ik $$ and hence $$ \sqrt{1+\gamma(1)^2}=e^{\tfrac{1}{2}\log(1+\gamma(1)^2)}=e^{\tfrac{1}{2}\log(1+\gamma(0)^2)+2\pi i k}=e^{\tfrac{1}{2}\log(\gamma(0)^2)}=\sqrt{1+\gamma(0)^2}. $$ Does this show that $\sqrt{1+z^2}$ may be analytically continued to all of $\mathbb{C}\setminus\{ix\in i\mathbb{R}\,:\,|x|<1\}$?

I know that if I could show that the analytic continuation is unique along any two curves with the same start and end points, I would be done. But this is a bit different. I feel like there just one more step of logic I need.

I'm also aware of a proof that $\sqrt{1+z^2}$ analytically continues to $\mathbb{C}\setminus\{ix\in i\mathbb{R}\,:\,|x|<1\}$, and while the proof makes sense and is elegant, I found it unintuitive, and I'm not likely to remember it. I would like to know if the result could be proved the way I have outlined instead.