What kinds of theories of real multiplication with one real constant are possible?

Via exponentiation, $(\Bbb R_{>0},\cdot)$ is the same as $(\Bbb R,+)$. Thereafter, $(\Bbb R,\cdot)$ is the same as the disjoint union of $\{0\}$ and $(\Bbb R_{>0},\cdot)$ and the negatives of $(\Bbb R_{>0},\cdot)$; here, the bijecton between positivies comes from multiplication with $-1$, which you saw is definable. Ultimately, everything we can say about $(\Bbb R,\cdot)$ can be expressed in terms of $(\Bbb R,+)$ this way. So indeed, for $(\Bbb R,\cdot,r)$, we have

• the same distinctions as fro $(\Bbb R,+,\ln r)$ if $r>0$, namely $r=1$ or $0<r\ne1$.
• the above cases transported to the negatives by multiplication with $-1$, i.e., $r=-1$ or $0>r\ne -1$
• the remaining case $r=0$

So the two cases from addition indeed translate into five cases for multiplication. If there were any additional case for multiplication, this would translate back to an additional case for addition.

Yes, you are correct.

We show something stronger: for any $a,b\in\mathbb{R}\setminus\{-1,0,1\}$ with the same sign there is an automorphism $\pi_{a,b}$ of $(\mathbb{R},\cdot)$ with $\pi(a)=b$. A fortiori $(\mathbb{R};\cdot, a)\equiv(\mathbb{R};\cdot,b)$.

Specifically, for $s\in\mathbb{R}\setminus\{0\}$ let $$f_s:\mathbb{R}\rightarrow\mathbb{R}: x\mapsto {\vert x\vert\over x}(\vert x\vert^s).$$ Each such $f_s$ is an automorphism of $(\mathbb{R};\cdot)$, the key points being $u^wv^w=(uv^w)$ and $u^w=v^w\implies u=v$ for $u,v\ge 0$ and $w\not=0$.

Now for $a,b\in\mathbb{R}\setminus\{-1,0,1\}$ with the same sign, taking the appropriate logarithm gives an $f_s$ with $f_s(a)=b$.