I'm trying to prove that the group $\text{SO}_2$ of real orthogonal $2 \times 2$ matrices with determinant $1$ is isomorphic to $S^1$. The isomorphism proposed by Benedict Gross in his lecture video is $A = \text{rot}(\theta) \mapsto e^{i \theta}$. Here is my attempt at proving this is an isomorphism. I believe I need to fix a basis ahead of time for this map to fully make sense (is that correct?), so I'm going to assume that each element of $\text{SO}_2$ is written relative to the standard basis of $\mathbb{R}^2$. I am taking for granted that each element of $\text{SO}_2$ can be regarded as a counterclockwise rotation of a vector in the plane by some angle $\theta$. I believe I need to restrict $\theta$ to $[0, 2\pi)$ to guarantee that the map is injective.
Define $\varphi: \text{SO}_2 \to S^1$ sending $A = \text{rot}(\theta) \mapsto z = e^{i \theta}$ for $\theta \in [0, 2\pi)$. I claim that $\varphi$ is an isomorphism. Let $z \in S^1$. Then $|z| = 1$, and hence $z = e^{i \theta}$ for some $\theta \in [0, 2\pi)$. Let $A \in \text{SO}_2$ be the matrix of a rotation by $\theta$. Then $\varphi(A) = e^{i \theta} = z$, so $\varphi$ is surjective. Now, suppose that $\varphi(A) = \varphi(B)$. Then $e^{i \theta} = e^{i \psi}$ for some $\theta, \psi \in [0, 2\pi]$. As $\theta$ was restricted to $[0, 2\pi)$, this forces $\theta = \psi$, so $\varphi$ is injective and hence bijective. Finally, let $A,B \in \text{SO}_2$, which represent counterclockwise rotations by angles $\theta$ and $\psi$, respectively. We then have \begin{align*} \varphi(AB) &= \varphi(\text{rot}(\theta) \circ \text{rot}(\psi)) \\ &= \varphi(\text{rot}(\theta + \psi)) \\ &= e^{i (\theta + \psi)} \\ &= e^{i \theta + i \psi} \\ &= e^{i \theta} e^{i \psi} \\ &= \varphi(A) \varphi(B), \end{align*} so $\varphi$ is a bijective homomorphism, so $\text{SO}_2 \cong S^1$, as required.
How does this look?
If you take for granted that the elements of $SO(2)$ can be regarded as counterclockwise plane rotations by some angle, you are essentially done. You have a surjective group homomorphism $$\rho : \mathbb R \to SO(2), \rho(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{pmatrix} = \text{ counterclockwise plane rotation by } \theta$$ with period $2\pi$ (i.e. $\rho(\theta + 2 \pi) = \rho(\theta)$). See for example Special Orthogonal Group $SO(2)$.
Your function $\varphi$ is then given as follows: For $A \in SO(2)$ pick $\theta \in \mathbb R$ such that $\rho(\theta) = A$ and set $\varphi(A) = e^{i\theta}$. By periodicity you may assume that $\theta \in [0,2\pi)$, but that is irrelevant. Actually you have to check that $\varphi(A)$ does not depend on the choice of $\theta$. But this is obvious because $\theta \mapsto e^{i\theta}$ also has period $2\pi$.
Here you have a gap in your proof: In general $\theta + \psi \notin[0,2\pi)$ for $\theta,\psi \in[0,2\pi)$.
A better approach is this. The above representation of $A \in SO(2)$ in the form $A = \rho(\theta)$ shows that $SO(2)$ is the set of all matrices $$A = \begin{pmatrix} a & -b\\ b & a \end{pmatrix}$$ such that $a^2 + b^2 = 1$. Now define $$\varphi : SO(2) \to S^1, \varphi\begin{pmatrix} a & -b\\ b & a \end{pmatrix} = a + ib. $$ It is easy to verify that $\varphi$ is a group isomorphism.