Dirac's equation necessitates the introduction of $4\times4$ complex matrices $\gamma^\mu$ ($\mu = 0, 1, 2, 3$) satisfying the Clifford algebra
$$\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu \nu}$$
where $\eta$ is the usual Minkowski metric. It's not hard to check that if $\Lambda = (\Lambda^\mu_\nu)$ is a Lorentz transformation, then the matrices
$$\widetilde{\gamma}^\mu = \Lambda^\mu_\nu \gamma^\nu$$
also satisfy the Clifford algebra. It then follows by the representation theory of Clifford algebras that there must be $S(\Lambda)$ such that
$$\Lambda^\mu_\nu \gamma^\nu = S(\Lambda) \gamma^\mu S(\Lambda)^{-1}$$
though it is not obvious that this $S$ can be picked to be a representation of the Lorentz group (or is it?) In any case we can search for a representation $S$ by figuring out how it should act on infinitesimal transformations. Thinking of $\Lambda = \Lambda(t)$, with $\Lambda(0) = I$, and defining
$$\omega^\mu_\nu = \frac{d}{dt} \Bigr|_{t=0} \Lambda^\mu_\nu(t)$$
the linearized equation is
$$\omega^\mu_\nu \gamma^\nu = [s(\omega), \gamma^\mu]$$
where $s$ is the sought-after representation of the Lorentz algebra. The usual formula given in physics texts works, and my question is not about that formula (though if you have some insight about it let me know).
My question is: supposing we do have a representation $s$ of the Lorentz algebra satisfying this linearized equation, how do we know that the original
$$\Lambda^\mu_\nu \gamma^\nu = S(\Lambda) \gamma^\mu S(\Lambda)^{-1}$$
holds, when defining $\Lambda = \exp(\omega)$ and $S(\Lambda) = \exp(s(\omega))$? This feels like an elementary question about matrix exponentials, but I've been looking at it for a while without ideas.
The book [1] giving good hints seems to be available only in German. On p. 286 they write $$ \psi\to\Lambda_S\,\psi\equiv\exp(-i\,\omega_{\mu\nu}S^{\mu\nu})\,\psi $$ where
$\omega_{\mu\nu}$ is related to the ordinary Lorentz transformaton ${\Lambda^\mu}_\nu$ by \begin{align}\tag{1.90} {\Lambda^\mu}_\nu&=\exp(M)\,,\\ \tag{1.92} {M^\rho}_\delta&=\omega^{\mu\nu}{{M_{\mu\nu}}^\rho}_\delta\,,\\ \tag{1.93} M_{\mu\nu\rho\delta}&=-g_{\mu\rho}g_{\nu\delta}+g_{\nu\rho}g_{\mu\delta}\,, \end{align} (equation numbering is from [1]), and
$\Lambda_S$ is your $S(\lambda)^{-1}\,,$
$$\tag{7.7} \boxed{{\Lambda^\mu}_\nu\gamma^\nu\equiv\Lambda^{-1}_S\,\gamma^\mu\Lambda_S\,,} $$ and
$$\tag{7.4} S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu] $$ are the spin-$\frac{1}{2}$-generators of the Lorentz transformatons.
I have not worked out any details myself yet and would be interested to hear from you what you think of this.
[1] L. Edelhäuser, A. Knochel, Tutorium Quantenfeldtheorie, Springer Spektrum 2016.
To complete the notation: $$\tag{1.54} g^{\mu\nu}=\left( \begin{matrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1 \end{matrix}\right)\,. $$