All solutions are "Hamel basis" solutions. Any solution of the functional equation is $\mathbb{Q}$-linear. Let $H$ be a Hamel basis. Given a solution $f$ of the functional equation, let $g(b)=f(b)$ for every $b\in H$, and extend by $\mathbb{Q}$-linearity. Then $f(x)=g(x)$ for all $x$.
There are lots of solutions because a Hamel basis has cardinality the cardinality $c$ of the continuum. A solution can assign arbitrary values to elements of the basis, and be extended to $\mathbb{R}$ by $\mathbb{Q}$-linearity. So there are $c^c$ solutions. There are only $c$ linear solutions, so "most" solutions of the functional equation are non-linear.
No you can't find a bijection on $\mathbb{R}$ with this property, the only function is $f\equiv 0$.
Indeed you have $f(0) = f(0\times x) = f(0) + f(x)$ so that $f(x) = 0$. This is true for all $x\in\mathbb{R}$.
If you exclude the $\{0\}$ again you cannot have a bijection since you have that $f$ must be a even function. Indeed $f(x)=\frac{1}{2}f(x^2)=f(-x)$.
However, forgetting about the bijection and looking just at $\mathbb{R}^+$ as a domain, it's true that $\log$ and its multiples are not the only possible functions. Indeed you may find everywhere discontinuous functions with the required property.
For instance you can build the function $f$ as follows. For each prime $p$ you fix $f(p) = k_p$, with $k_p$ any arbitrary number. Then for any $q\in\mathbb{Q}$ you have that $f(p^q)=qf(p)=qk_p$. The value of $f$ is fixed for all the $x$ which can be written as finite product of this terms, i.e. such that there exist a finite family of prime numbers $\{p_i\}_{i=1\dots N}$ and rational coefficients $\{q_i\}_{i=1\dots N}$ and
$$x = \prod_{i=1}^N {p_i}^{q_i}\,.$$
You have indeed $f(x) = \sum_{i=1}^N q_i k_{p_i}$. You can show that each of these $x$ has a unique representation in this form, so that $f$ is well defined for them.
For the $x$ which are not representable as such a finite product you define $f(x) = 0$. Then $f$ is well defined on $\mathbb{R}^+$ and your property holds.
Edit: To answer your Edit...
All the continuous functions on $\mathbb{R}^+$ which respect the property are in the form $\alpha \log$ for some real $\alpha$. Actually, if $f$ is continuous at $x=1$ then it's in the form $\alpha\log$.
Indeed assume that f is continuous at $1$.
First notice that $f(1) = 0$ since $f(x) = f(1\times x) = f(1) + f(x)$.
Now you can prove that $f$ is continuous everywhere since
$$\lim_{h\to 0} f(x+h) = \lim_{h\to 0}f(x(1+h/x)) = f(x)+\lim_{h\to 0}f(1+h/x) = f(x)+f(1) = f(x)\,.$$
Now you know that for any rational $q$, $f(e^q) = qf(e)$. By continuity you have that for any $r\in\mathbb{R}$
$$f(e^r) = r f(e)\,.$$
So $$f(x) = f(e^{\log x}) = f(e)\log x = \alpha \log x\,,$$ letting $\alpha=f(e)$.
For the sake of completeness, I don't think it is of much sense to speak of the property for a function defined on a generic $D\subset \mathbb{R}^+$. For instance for a open interval $D=(10,11)$, you have no $x_1,x_2\in(10,11)$ such that $x_1\cdot x_2 \in (10,11)$, which means that any function satisfies the property. However, every time that you have a neighbourhood of $1$ in $D$, and $f$ is continuous at $1$, then in each open connected component $f$ must be in the form $x\mapsto k+\alpha \log x$, with $k$ which can possibly vary in different connected components.
Best Answer
Suppose that $f(x^t) = t f(x)$ for all $x>0$ and all $t\in\Bbb R$. In particular, $f(x) = f(e^{\log x}) = (\log x)f(e)$ for all $x>0$. Therefore $f(x)$ must be a logarithm function, namely $f(x) = \log_bx$ with $b=e^{1/f(e)}$. (One exception: if $f(e)=0$ then $f(x)=0$ identically.)