Levi factors are noncanonical but conjugate

Projection to $\mathfrak{g}\mathrm{/}\mathrm{rad}(\mathfrak{g})$ restricts to an injective map on a Levi factor because the intersection is zero, and to a surjective map because the two subspaces sum to $\mathfrak{g}$. It is a Lie homomorphism, so every Levi factor is isomorphic to the quotient.

By [L1], any two factors are related by the stated finite product. Each $\mathrm{ad}x$ is nilpotent in the proof of [L1], so its exponential is unipotent and is inner in the Lie-algebra sense. When the radical is zero, the sole factor is $\mathfrak{g}$ and the empty product suffices.

To witness literal nonuniqueness, let $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(k)$ act in the standard way on $V=k^2$ and put $\mathfrak{g}=\mathfrak{s}⋉V$. For the usual basis $h,e,f$ of $\mathfrak{s}$, direct calculation gives $K(h,h)=8$, $K(e,f)=K(f,e)=4$, and all other basis pairings zero; the Killing matrix has determinant $-128\mathrm{\ne }0$, so [L4] makes $\mathfrak{s}$ semisimple. By [L2], $V$ is an abelian ideal and $\mathfrak{g}\mathrm{/}V\cong \mathfrak{s}$. Hence [L3] gives $V\subseteq \mathrm{rad}(\mathfrak{g})$, while the image of the radical in the semisimple quotient is a solvable ideal and is zero; thus the radical is exactly $V$ and ${\mathfrak{s}}_0=\mathfrak{s}\oplus 0$ is a Levi factor. Take $v=(1,0{)}^T$ and $x=f$, so $xv=(0,1{)}^T\mathrm{\ne }0$. Formula [L2] gives $({\mathrm{ad}}_{(0,v)}{)}^2=0$ and $\exp ({\mathrm{ad}}_{(0,v)})(x,0)=(x,-xv)\mathrm{\notin }{\mathfrak{s}}_0$. Therefore ${\mathfrak{s}}_1=\exp ({\mathrm{ad}}_{(0,v)})({\mathfrak{s}}_0)$ is a distinct Levi factor, proving the first sentence rather than merely referring to a later example.