The usual way to show that the topological groups $(\mathbb{R}_+^*,\cdot)$ and $(\mathbb{R},+)$ are isomorphic is to consider, for instance, the function $\operatorname{ln}\colon \mathbb{R}_+^*\to\mathbb{R}$ which maps $x\mapsto \int_1^x\frac{1}{t}dt$. On the other hand
every topological group $G$ homeomorphic to $\mathbb{R}$ must be isomorphic to $(\mathbb{R},+)$,
which it follows from Theorem 2 in Bourbaki, for instance.
Since every Hausdorff locally compact topological group $G$ has a Haar integral, i.e., a nontrivial left invariant positive linear functional $\mu\colon C_c(G)\to\mathbb{R}$, I would like to know if there is an adaptation of the logarithm construction that allows one to "simplifies" Bourbaki's proof of the quoted result above $-$ or if Haar integrals can be helpful in any other way in this context. References where this is done are welcome as well.
Sure; the approach to take becomes clear when you notice that $\frac{1}{t}dt$ is a Haar measure on $\mathbb{R}_+$. So suppose we have a topological group structure $*$ on $\mathbb{R}$ with left Haar measure $\mu$; we wish to show this structure is isomorphic as a topological group to the usual one. We may assume the identity element of the group structure is $0$. Now define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\int_0^x d\mu$ (that is, $f(x)=\mu([0,x])$ if $x\geq 0$ and $f(x)=-\mu([x,0])$ if $x<0$); I claim this is the desired isomorphism. It is easy to see that this $f$ is strictly increasing and continuous, so it suffices to show that $f$ is a homomorphism from $(\mathbb{R},*)$ to $(\mathbb{R},+)$. To show this, just observe that $$f(x*y)=\int_0^{x*y}d\mu=\int_0^xd\mu+\int_x^{x*y}d\mu=\int_0^xd\mu+\int_0^yd\mu=f(x)+f(y)$$ where the third equality uses the left-invariance of $\mu$.