This is related to an earlier question of mine. I would like an argument or a reference (or a missing hypothesis if needed) for the following.

Let $\mathbb{F}\langle\langle \alpha\rangle\rangle$ and $\mathbb{F}\langle\langle \beta\rangle\rangle$ be the algebras of formal power series in the given sets of non-commuting indeterminates over a field $\mathbb{F}$ of characteristic zero. Let $\mathfrak{g}$ and $\mathfrak{h}$ be the respective Lie subalgebras with zero constant term. Let $G$ and $H$ be the subgroups of the groups of units of the respective algebras consisting of the elements with constant term equal to 1. (So $\log$ maps $G$ bijectively onto $\mathfrak{g}$ and similarly for $H$.) Let $\Phi:G\to H$ be a homomorphism, and set $\phi=\log\Phi\exp$. Assume that $\phi(aX)=a\phi(X)$ for all $a$ and $X$. Then $\phi(X+Y)=\phi(X)+\phi(Y)$.

Let's recall the set-up in the classical case of matrix Lie groups and Lie algebras. Under suitable hypotheses there is a 1-1 correspondence between Lie group homomorphisms $\Phi:G\to H$ and Lie algebra homomorphisms $\phi:\mathfrak{g}\to\mathfrak{h}$ where $\mathfrak{g}$ is the Lie algebra corresponding to $G$, and similarly for $\mathfrak{h}$: they are related via $\exp\phi=\Phi\exp$. The proof that a Lie algebra homomorphism $\phi$ gives rise to a Lie group homomorphism $\Phi$ is fairly neat. A key step is to show that $\log(\exp(\phi(X))\exp(\phi(Y)))=\phi(\log(\exp(X)\exp(Y)))$, which is a consequence of the Baker-Campbell-Hausdorff formula.

To show that $\Phi$ gives rise to a suitable $\phi$, the key step this time is to show that $\phi$ is additive which reduces to the identity

$\Phi\exp(\log g + \log h)=\exp(\log\Phi(g)+\log\Phi(h))$,

but this seems to be trickier to prove in the situation I'm interested in. Even if one inverts the BCH formula (see this question and answer), the required identity is not obvious to me. (In the notation of that answer, we would need the identity $\Phi A(g-1,h-1) = A(\Phi(g)-1,\Phi(h)-1)$.)

The proof in Brian C. Hall's book to handle the classical case uses $\left(e^{\frac{X}{m}}e^{\frac{Y}{m}}\right)^m$ as an approximation to $e^{X+Y}$ and then involves taking $\lim_{m\to\infty}$, but unfortunately taking such limits doesn't seem to be valid in my situation. This could be remedied by assuming $\mathbb{F}$ is an ordered field and that $\Phi$ preserves limits in an appropriate sense, but I would like to avoid making extra assumptions if possible.