I've tried to build $SO (3)$ starting from $\mathfrak{su} (2)$ and the Pauli matrices. I easily computed the elements in $\mathfrak{so} (3)$ by applying the homomorphic map $$ \mathcal{L} \to \mathcal{L'}, \quad a \mapsto \Psi (a)_{jk} = \dfrac{1}{2} \operatorname{Tr} \bigl( \sigma_j [a, \sigma_k] \bigr), $$
where $\sigma_i$ are the Pauli matrices and $\mathfrak{su}(2)$ generators. But when trying to apply an exponential map s.t. $$ \theta (\Psi) = e^{\Psi (a)} = \sum_{k=0}^{\infty} \dfrac{1}{k!} \left(\Psi (a) \right)^k \in SO (3), $$ I couldn't succeed.
I'm only a physicist, so I don't know the level of rigor you are after. You constructed 3×3 antisymmetric, hence traceless matrices $\Psi(a)=-\Psi(a)^T$, depending on the 3-vector parameters $\vec a$, implicit in your $a=\vec a\cdot \vec \sigma$.
The exponential of $\Psi(a)$ is then an orthogonal matrix O, $$ O(a) O(a)^T =e^{\Psi(a)} e^{-\Psi(a)}= {\mathbb I}; $$ but O is also unimodular, as $$ \det O(a)=e^{\operatorname{Tr}(\log e^{O})}= e^{\operatorname{Tr} O}=1. $$ Don't you define SO(3) as the group represented by orthogonal unimodular 3×3 matrices?