Below I will give a proof of the title's question. I was not sure if I had to prove that each of the used bijections is a homomorphism as well, but I just wanted to also go through (perhaps trivial) steps. Any tips or corrections are greatly appreciated!
Let $\mathbb{R}_{>1}=\{x\in\mathbb{R}:x>1\}$ (analogous for $\mathbb{R}_{>0}$). We define $\circ$ by $a\circ b = a^{\log b}$, $+$ and $*$ as "usual".
Prove that $(\mathbb{R},+) \cong (\mathbb{R}_{>1},\circ)$
Proof:
We will show that $(\mathbb{R},+)\cong(\mathbb{R}_{>0},*)$ and $(\mathbb{R}_{>0},*)\cong (\mathbb{R}_{>1},\circ)$.
First, we can look at the bijection $f:(\mathbb{R},+)\to (\mathbb{R}_{>0},*)$ defined by $f(x)=e^x$.
$f(a+b)=e^{a+b}=e^a*e^b=f(a)*f(b)$ so $f$ is a group homomorphism.
$R(f)=\{y\in\mathbb{R}_{>0}:e^x=y\}$
We know that $\forall_{x\in\mathbb{R}}:0<e^x<\infty$ such that $R(f)=\mathbb{R}_{>0}$. Thus: $f$ is surjective.
$N(f)=\{x\in\mathbb{R}:e^x=1\}$
We know that $\forall_{x\in\mathbb{R}}:e^x=1\iff x=0$ such that $N(f)=\{0\}$. Thus: $f$ is injective.
Together this gives that $f$ is a bijection, so $(\mathbb{R},+)\cong(\mathbb{R}_{>0},*)$.
Now, we can look at the bijection $g:(\mathbb{R}_{>0},*)\to (\mathbb{R}_{>1},\circ)$ defined by $g(x)=e^x$.
$g(a*b)=e^{ab}=(e^a)^b=(e^a)^{\log e^b}=e^a\circ e^b = g(a)\circ g(b)$ so $g$ is a group homomorphism.
$R(g)=\{y\in\mathbb{R}_{>1}:e^x=y\}$
We know that $\forall_{x\in\mathbb{R}_{>0}}:1<e^x<\infty$ such that $R(g)=\mathbb{R}_{>1}$. Thus: $g$ is surjective.
$N(g)=\{x\in\mathbb{R}_{>0}:e^x=e\}$
We know that $\forall_{x\in\mathbb{R}_{>0}}:e^x=e\iff x=1$ such that $N(g)=\{1\}$. Thus: $g$ is injective.
Note that the identity element of $(\mathbb{R}_{>1},\circ)$ is $e$, as $\forall_a: a\circ e=a^{\log e} = a = e^{\log a} = e\circ a.$
Together this gives that $g$ is a bijection, so $(\mathbb{R}_{>0},*)\cong (\mathbb{R}_{>1},\circ)$.
By the transitivity of $\cong$, we now have that $(\mathbb{R},+) \cong (\mathbb{R}_{>1},\circ)$ $\tag*{$\Box$}$