Okay, based on Max0815's suggested link, along with a handful of Wikipedia pages, I came up with a somewhat self-contained solution to this problem for anyone unfamiliar with ring and/or field theory.
The main idea is to use the Lagrange Inversion formula, which I frankly still don't fully understand.
But let me show only its application for this problem.
Consider any function $f$ that is analytic on some deleted neighborhood of $z=0$.
It can be expanded as the Laurent series
$$ f(z) = \sum_{n = -\infty} ^ {\infty} a_n z^n.$$
Let us denote the coefficient of $x^n$ as $\left[ z^n \right] f(z)$, which is determined uniquely since $f$ is analytic.
We clearly see that for any integers $n$ and $k$, $$\left[ z^n \right] f(z) = \left[ z^k \right] \left( z^{k - n} f(z) \right). $$
The first goal of the proof is to show that for any integers $k$ and $n$,
$$ k \cdot \left[ z^k \right] \left( \sin^{-1} z \right)^n = n \cdot [z^{-n}] \left( \sin z \right)^{-k}. $$
For the record, since $\sin^{-1}$ is multivalued, we choose the branch of $Re\sin^{-1} z \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right]$. (Not that it matters for our purposes...)
From the above discussion, we have $$ k \cdot \left[ z^k \right] \left( \sin^{-1} z \right)^n = k\cdot \left[ z^{-1} \right] \left( z^{-k-1} \left( \sin^{-1} z \right)^n \right). $$
Let $g(z):=z^{-k-1} \left( \sin^{-1} z \right)^n$.
It clearly has a convergent Laurent series near $z=0$, which may or may not include a $z^{-1}$ term.
Let $g_{-1}:=\left[ z^{-1} \right]g(z)$.
Then, the Laurent series for $g(z)-\frac{g_{-1}}{z}$ can be termwise integrated to yield a new function $G(z)$, analytic on a deleted neighborhood of $0$.
Rewriting $g(z)=\frac{g_{-1}}{z} + G'(z)$, we see that
$$ g(\sin z) \cdot \cos z = g_{-1} \cot z + (G(\sin z))'. $$
It is straightforward to calculate that $\left[ z^{-1} \right] \cot z = 1$.
As for the latter, notice that the derivative of no integer power of $z$ yields a $z^{-1}$ term. Hence, we have
$$ \left[ z^{-1} \right] (G(\sin z))' = 0 ~\Rightarrow~ \left[ z^{-1} \right] \left( g(\sin z) \cos z \right) = g_{-1} = \left[ z^{-1} \right] g(z)$$
$$
\begin{equation}
\begin{split}
\therefore k\cdot \left[ z^{-1} \right] \left( z^{-k-1} \left( \sin^{-1} z \right)^n \right)
& = k\cdot \left[ z^{-1} \right] \left( (\sin z)^{-k-1} z^n \right) \\
& = -\left[ z^{-1} \right] \left( z^n \left( \left( \sin z \right)^{-k} \right)' \right)
\end{split}
\end{equation}
$$
Recall from before that the derivative of any analytic function cannot have any $z^{-1}$ terms.
Then, it follows that
$$ \left[ z^{-1} \right] \left( z^n \left( \sin z \right)^{-k} \right)' = 0. $$
Therefore, by chain rule, we have
$$
\begin{equation}
\begin{split}
-\left[ z^{-1} \right] \left( z^n \left( \left( \sin z \right)^{-k} \right)' \right)
& = \left[ z^{-1} \right] \left( \left( z^n \right)' \left( \sin z \right)^{-k} \right) \\
& = \left[ z^{-1} \right] \left( nz^{n-1} \cdot \left( \sin z \right)^{-k} \right) \\
& = n \cdot [z^{-n}] \left( \sin z \right)^{-k}
\end{split}
\end{equation}
$$
Therefore, connecting the results, we have the desired inversion formula
$$ k \cdot \left[ z^k \right] \left( \sin^{-1} z \right)^n = n \cdot [z^{-n}] \left( \sin z \right)^{-k}. $$
Notice that, by definition, $ Res_{z = 0} f(z) = \left[ z^{-1} \right] f(z) $.
Thus, the residue that we want is
$$
\begin{equation}
\begin{split}
Res_{z=0} e^{\frac{1}{\sin z}}
& = \left[ z^{-1} \right] e^{\frac{1}{\sin z}} \\
& = \left[ z^{-1} \right] \sum_{k=0}^{\infty} \frac{1}{k!} \cdot \left( \frac{1}{\sin z} \right)^k \\
& = \sum_{k=1}^{\infty} \frac{1}{k!} \cdot 1 \cdot \left[ z^{-1} \right] \left( \sin z \right)^{-k} \\
& = \sum_{k=1}^{\infty} \frac{1}{k!} \cdot k \cdot \left[ z^k \right] \left( \sin^{-1} z \right)^{1} \\
& = \sum_{k=1}^{\infty} \frac{1}{(k-1)!}\left[ z^k \right] \sin^{-1} z
\end{split}
\end{equation}
$$
Noticing $\left( \sin^{-1} z \right)' = \left( 1 - z^2 \right)^{-1/2}$ and termwise integrating the Taylor series of the right hand side, we get
$$
\begin{equation}
\begin{split}
\sin^{-1} z
& = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \begin{pmatrix} -1/2 \\ k \end{pmatrix} z^{2k+1} \\
& = \sum_{k=0}^\infty \frac{(2k)!}{\left( 2^k \cdot k! \right)^2 \cdot (2k+1)} z^{2k+1}
\end{split}
\end{equation}
$$
Finally, substituting it into our residue result, we get the final answer
$$
\begin{equation}
\begin{split}
Res_{z=0} e^{\frac{1}{\sin z}}
& = \sum_{k=0}^{\infty} \frac{1}{((2k+1)-1)!} \frac{(2k)!}{\left( 2^k \cdot k! \right)^2 \cdot (2k+1)} \\
& = \sum_{k=0}^{\infty} \frac{1}{\left( 2^k \cdot k! \right)^2 \cdot (2k+1)}
\end{split}
\end{equation}
$$
Any details I glossed over or criticisms are welcomed.
Thank you again to Max.
