[proofplan]
We use exponential coordinates near the identity in both Lie groups. In those coordinates, the local multiplication law is described by the Baker--Campbell--Hausdorff series. Since $\Phi$ preserves Lie brackets, it carries every iterated bracket in the BCH series for $\mathfrak g$ to the corresponding iterated bracket in $\mathfrak h$. Therefore the coordinate map $X\mapsto \Phi X$ intertwines the local group laws, and conjugating back by the two exponential maps gives the required local diffeomorphism.
[/proofplan]
[step:Choose exponential coordinates compatible with $\Phi$]
Let
\begin{align*}
\exp_G:\mathfrak g\to G
\end{align*}
and
\begin{align*}
\exp_H:\mathfrak h\to H
\end{align*}
denote the Lie group exponential maps. Since $G$ and $H$ are finite-dimensional real Lie groups, the standard local diffeomorphism theorem for the Lie group exponential map applies to $\exp_G$ and $\exp_H$ at $0$. Hence there exist open neighbourhoods $A_0\subset\mathfrak g$ of $0$ and $B_0\subset\mathfrak h$ of $0$ such that
\begin{align*}
\exp_G|_{A_0}:A_0\to \exp_G(A_0)
\end{align*}
and
\begin{align*}
\exp_H|_{B_0}:B_0\to \exp_H(B_0)
\end{align*}
are diffeomorphisms onto open neighbourhoods of $e_G$ and $e_H$, respectively.
Since $\Phi:\mathfrak g\to\mathfrak h$ is a linear isomorphism and $B_0$ is open with $0\in B_0$, the set
\begin{align*}
A_1:=A_0\cap \Phi^{-1}(B_0)
\end{align*}
is an open neighbourhood of $0$ in $\mathfrak g$. Shrinking $A_1$ if necessary, choose an open neighbourhood $A\subset A_1$ of $0$ such that the BCH product in $\mathfrak g$ is defined on the required local domain inside $A$, and such that the corresponding BCH product in $\mathfrak h$ is defined on $\Phi(A)\times\Phi(A)$. This shrinking is possible by the local BCH theorem for finite-dimensional Lie algebras, applied to $\mathfrak g$ and $\mathfrak h$; see [citetheorem:8798].
Define
\begin{align*}
U:=\exp_G(A)
\end{align*}
and
\begin{align*}
V:=\exp_H(\Phi(A)).
\end{align*}
Then $U$ is an open neighbourhood of $e_G$ in $G$, and $V$ is an open neighbourhood of $e_H$ in $H$.
[guided]
The first task is to put both groups into coordinates where their local products can be compared. The natural coordinates near the identity are exponential coordinates. Let
\begin{align*}
\exp_G:\mathfrak g\to G
\end{align*}
and
\begin{align*}
\exp_H:\mathfrak h\to H
\end{align*}
be the exponential maps of the Lie groups $G$ and $H$.
We use the standard local theorem that the exponential map is a diffeomorphism from a neighbourhood of $0$ in the [Lie algebra](/page/Lie%20Algebra) onto a neighbourhood of the identity in the Lie group. Thus there are open neighbourhoods $A_0\subset\mathfrak g$ and $B_0\subset\mathfrak h$ of $0$ such that
\begin{align*}
\exp_G|_{A_0}:A_0\to \exp_G(A_0)
\end{align*}
and
\begin{align*}
\exp_H|_{B_0}:B_0\to \exp_H(B_0)
\end{align*}
are diffeomorphisms onto open identity neighbourhoods. This is the coordinate chart input for the proof.
We also need the [linear map](/page/Linear%20Map) $\Phi$ to carry the chosen domain in $\mathfrak g$ into the chosen domain in $\mathfrak h$. Since $\Phi$ is continuous and invertible, the set
\begin{align*}
A_1:=A_0\cap \Phi^{-1}(B_0)
\end{align*}
is an open neighbourhood of $0$ in $\mathfrak g$. If $X\in A_1$, then $X\in A_0$ and $\Phi X\in B_0$, so both $\exp_G X$ and $\exp_H(\Phi X)$ are represented in the chosen exponential coordinate charts.
Finally, we shrink $A_1$ to an open neighbourhood $A$ of $0$ on which the BCH multiplication law is valid. The relevant local BCH result is [citetheorem:8798]: for each finite-dimensional Lie algebra, after shrinking to a sufficiently small neighbourhood of $0$, the expression $\operatorname{BCH}(X,Y)$ is defined whenever $X$, $Y$, and their product are kept in that neighbourhood, and it gives the local Lie group law. We apply this theorem to the finite-dimensional Lie algebras $\mathfrak g$ and $\mathfrak h$. Because $\Phi$ is a linear homeomorphism, we choose $A\subset A_1$ small enough that whenever $X,Y\in A$ and $\operatorname{BCH}_{\mathfrak g}(X,Y)\in A$, the pair $(\Phi X,\Phi Y)$ lies in the BCH domain for $\mathfrak h$ and $\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y)\in \Phi(A)$.
Now set
\begin{align*}
U:=\exp_G(A)
\end{align*}
and
\begin{align*}
V:=\exp_H(\Phi(A)).
\end{align*}
Because the restricted exponential maps are diffeomorphisms on these neighbourhoods, $U$ and $V$ are open neighbourhoods of $e_G$ and $e_H$.
[/guided]
[/step]
[step:Define the candidate local diffeomorphism]
Define
\begin{align*}
F:U\to V,\quad F(\exp_G X):=\exp_H(\Phi X)\text{ for }X\in A.
\end{align*}
This is well-defined because $\exp_G|_A$ is injective. Equivalently,
\begin{align*}
F=\exp_H\circ \Phi\circ (\exp_G|_A)^{-1}.
\end{align*}
Since $\exp_G|_A$, $\Phi$, and $\exp_H|_{\Phi(A)}$ are diffeomorphisms onto their images, $F$ is a diffeomorphism. Its inverse is
\begin{align*}
F^{-1}=\exp_G\circ \Phi^{-1}\circ (\exp_H|_{\Phi(A)})^{-1}.
\end{align*}
[guided]
The intended map is forced by the requirement that its differential at the identity be $\Phi$: in exponential coordinates it must send the coordinate vector $X\in A\subset\mathfrak g$ to the coordinate vector $\Phi X\in \Phi(A)\subset\mathfrak h$. Thus define
\begin{align*}
F:U\to V,\quad F(\exp_G X):=\exp_H(\Phi X)\text{ for }X\in A.
\end{align*}
This formula is well-defined because $\exp_G|_A:A\to U$ is injective, so each element of $U$ has a unique exponential coordinate $X\in A$. In composition form,
\begin{align*}
F=\exp_H\circ \Phi\circ (\exp_G|_A)^{-1}.
\end{align*}
Each factor in this composition is a diffeomorphism onto its image: $\exp_G|_A$ and $\exp_H|_{\Phi(A)}$ by the chosen exponential coordinates, and $\Phi$ because it is a linear isomorphism between finite-dimensional real vector spaces. Therefore $F$ is a diffeomorphism from $U$ onto $V$. Its inverse is explicitly
\begin{align*}
F^{-1}=\exp_G\circ \Phi^{-1}\circ (\exp_H|_{\Phi(A)})^{-1}.
\end{align*}
[/guided]
[/step]
[step:Show that $\Phi$ preserves the BCH product]
Let $\operatorname{BCH}_{\mathfrak g}$ denote the BCH series formed using the bracket on $\mathfrak g$, and let $\operatorname{BCH}_{\mathfrak h}$ denote the BCH series formed using the bracket on $\mathfrak h$. Since $\Phi$ is a Lie algebra homomorphism, for every $X,Y\in\mathfrak g$,
\begin{align*}
\Phi([X,Y])=[\Phi X,\Phi Y].
\end{align*}
By induction on the bracket depth, $\Phi$ carries every iterated bracket in $X$ and $Y$ to the corresponding iterated bracket in $\Phi X$ and $\Phi Y$. Since the BCH series is a universal Lie series in iterated brackets, on the chosen BCH convergence neighbourhood from [citetheorem:8798] we have
\begin{align*}
\Phi(\operatorname{BCH}_{\mathfrak g}(X,Y))=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y).
\end{align*}
[guided]
We now verify that the coordinate change $X\mapsto \Phi X$ respects the BCH products. Let $\operatorname{BCH}_{\mathfrak g}$ be the BCH series computed with the Lie bracket of $\mathfrak g$, and let $\operatorname{BCH}_{\mathfrak h}$ be the BCH series computed with the Lie bracket of $\mathfrak h$. The theorem [citetheorem:8798] gives the local BCH domains after the shrinking performed above, so the following expressions are defined for the pairs used in the multiplication argument.
Because $\Phi:\mathfrak g\to\mathfrak h$ is a Lie algebra homomorphism, for every $X,Y\in\mathfrak g$,
\begin{align*}
\Phi([X,Y])=[\Phi X,\Phi Y].
\end{align*}
Applying this identity repeatedly shows, by induction on bracket depth, that every iterated bracket in $X$ and $Y$ is carried by $\Phi$ to the same iterated bracket in $\Phi X$ and $\Phi Y$. For example, the bracket $[X,[X,Y]]$ is sent to
\begin{align*}
\Phi([X,[X,Y]])=[\Phi X,\Phi([X,Y])]=[\Phi X,[\Phi X,\Phi Y]].
\end{align*}
The BCH series is universal: its terms are fixed rational linear combinations of these iterated brackets. Therefore applying $\Phi$ term by term to the BCH series in $\mathfrak g$ gives exactly the BCH series in $\mathfrak h$ evaluated at $\Phi X$ and $\Phi Y$. Hence, on the chosen convergence neighbourhood,
\begin{align*}
\Phi(\operatorname{BCH}_{\mathfrak g}(X,Y))=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y).
\end{align*}
[/guided]
[/step]
[step:Translate the BCH identity back to group multiplication]
Let $g,h\in U$ and suppose $gh\in U$. Since $\exp_G|_A:A\to U$ is a diffeomorphism, there are unique $X,Y,Z\in A$ such that
\begin{align*}
g=\exp_G X,\quad h=\exp_G Y,\quad gh=\exp_G Z.
\end{align*}
By the BCH description of the local multiplication law in exponential coordinates from [citetheorem:8798], and by the choice of $A$, the condition $gh\in U$ implies
\begin{align*}
Z=\operatorname{BCH}_{\mathfrak g}(X,Y).
\end{align*}
Therefore
\begin{align*}
F(gh)=F(\exp_G Z)=\exp_H(\Phi Z).
\end{align*}
Using the BCH preservation identity from the previous step gives
\begin{align*}
\Phi Z=\operatorname{BCH}_{\mathfrak h}(\Phi X,\Phi Y).
\end{align*}
Applying the BCH multiplication formula in $H$ gives
\begin{align*}
\exp_H(\Phi Z)=\exp_H(\Phi X)\exp_H(\Phi Y).
\end{align*}
By the definition of $F$,
\begin{align*}
\exp_H(\Phi X)\exp_H(\Phi Y)=F(g)F(h).
\end{align*}
Combining these equalities yields
\begin{align*}
F(gh)=F(g)F(h).
\end{align*}
[/step]
[step:Conclude that $F$ is a local Lie group isomorphism]
We have constructed open neighbourhoods $U\subset G$ of $e_G$ and $V\subset H$ of $e_H$, and a diffeomorphism $F:U\to V$. The preceding step proves that for all $g,h\in U$ satisfying $gh\in U$,
\begin{align*}
F(gh)=F(g)F(h).
\end{align*}
Thus $F$ is the required local product-preserving diffeomorphism.
[/step]