Based on your question, you are using a very limited form of Cauchy's theorem. The general theorem (even when restricted to just curves) does not require the domain to be simply-connected, nor that the curve be simple. It only requires that the winding number of the curve around any point not in the domain be $0$.
However, this must be done with the tools you have, not with the tools you wish you had, so we'll stick with these restrictions:
- Cauchy's theorem only applies to simply-connected domains and to simple closed curves (but at least, $L$ is a simple closed curve).
- No Cauchy Integral Formula. (Residues are a consequence of Cauchy's integral formula, so they are out too.)
First, choose a point $p$ of minimum magnitude on $L$, and for $\epsilon > 0$, another point $q_\epsilon$ on $L$ within $\epsilon$ of $p$. Let $r$ be such that $1 < r < |p|$, and drop parallel line segments $\ell, \ell_\epsilon$ down from $p$ and $q_\epsilon$ respectively to the circle $C$ of radius $r$ about the origin. Then form the curve $L'$ that
- Follows $L$ from $p$ the long way around to $q_\epsilon$,
- traverses the line $\ell_\epsilon$ from $q_\epsilon$ down to $C$,
- traverses $C$ in the opposite direction around till it intersects with $\ell$.
- traverses $\ell$ back to $p$ to close.
Then $L'$ is a simple closed curve. Further, we can split the domain $U$ by a line raising up from the unit circle midway between $\ell$ and $\ell_\epsilon$ until it crosses $L$. After that, we can continue this as a curve that follows $L$ closely without touching until it reaches a point where it can extend to $\infty$ without intersecting $L$. When this curve is removed from $U$, the remainder is simply connected. Thus we can apply Cauchy's theorem to $L'$ to say that $\oint_{L'} f(z)dz = 0$
As you let $\epsilon \to 0$, $q_\epsilon \to p$ and $\ell_\epsilon \to \ell$, so $$0 = \oint_{L'} f(z)dz \to \oint_L f(z)dz + \int_{\ell} f(z)dz - \oint_C f(z)dz - \int_{\ell} f(z)dz$$
From which it follows that $$\oint_L f(z)dz = \oint_C f(z)dz$$
So all that is left is to show that $\oint_C f(z)dz = 0$. But if $C'$ is a circle about the origin of higher radius than $C$, by letting $C'$ take the place of $L$ in the result just shown, they have the same integral. Thus $\oint_C f(z)dz$ does not depend on the radius $r$ of $C$. Now $$\oint_C f(z)dz = i\int_0^{2\pi} \frac{rd\theta}{e^{-i\theta} + r^2e^{i\theta}}$$
As $r \to \infty$, this converges to $0$. But since it is constant, it had to be $0$ all along, which completes the proof (except for cleaning up all the handwaving).
