= Field invariants
Let $\mathbf u$ be the boost <velocity>, and decompose the fields into components parallel and perpendicular to it. The parallel components are unchanged. The perpendicular components obey
$$
\mathbf E'_\perp=\gamma(\mathbf E_\perp+\mathbf u\times\mathbf B_\perp),\qquad
\mathbf B'_\perp=\gamma(\mathbf B_\perp-\mathbf u\times\mathbf E_\perp/c^2),\qquad
\gamma=(1-u^2/c^2)^{-1/2}.
$$
All vectors inside these expressions are perpendicular to $\mathbf u$. Thus $|\mathbf u\times\mathbf B_\perp|^2=u^2B_\perp^2$, and similarly for $\mathbf E_\perp$. In the difference of squares, the mixed terms cancel because
$$
\mathbf E_\perp\cdot(\mathbf u\times\mathbf B_\perp)
=-\mathbf B_\perp\cdot(\mathbf u\times\mathbf E_\perp).
$$
It follows that
$$
\begin{aligned}
E_\perp'^2-c^2B_\perp'^2
&=\gamma^2\left[(1-u^2/c^2)E_\perp^2-(c^2-u^2)B_\perp^2\right]\\
&=E_\perp^2-c^2B_\perp^2.
\end{aligned}
$$
Adding the unchanged parallel components proves
$$
\boxed{E'^2-c^2B'^2=E^2-c^2B^2.}
$$
For the mixed <dot product>, expand directly. The terms $\mathbf E_\perp\cdot(\mathbf u\times\mathbf E_\perp)$ and $(\mathbf u\times\mathbf B_\perp)\cdot\mathbf B_\perp$ vanish. The remaining cross-product identity is
$$
(\mathbf u\times\mathbf B_\perp)\cdot(\mathbf u\times\mathbf E_\perp)
=u^2\mathbf E_\perp\cdot\mathbf B_\perp.
$$
Hence
$$
\mathbf E'_\perp\cdot\mathbf B'_\perp
=\gamma^2(1-u^2/c^2)\mathbf E_\perp\cdot\mathbf B_\perp
=\mathbf E_\perp\cdot\mathbf B_\perp.
$$
Parallel-perpendicular mixed <dot products> are zero in both frames, so
$$
\boxed{\mathbf E'\cdot\mathbf B'=\mathbf E\cdot\mathbf B.}
$$
Rotations also preserve these two expressions, completing the proof for proper <Lorentz transformations>. The second expression is a pseudoscalar under spatial parity, but is invariant under the boosts at issue here.
These results also follow from <tensor contractions>. In the convention used below, $F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2)$ and $F_{\mu\nu}\widetilde F^{\mu\nu}=-4\mathbf E\cdot\mathbf B/c$. The numerical sign of the latter depends on the chosen orientation of the <dual electromagnetic field tensor>, while its invariance under proper <Lorentz transformations> does not.
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