This is not an answer but an attempt to clarify the question.
In the category of right $G$-spaces (with weak equivalences being the maps that as maps of spaces are weak equivalences) let us single out those objects $X$ for which there is a weak equivalence $G\to X$ (where $G$ has the usual right action). In other words, those such that for some $x$ the map $g\mapsto xg$ is a weak equivalence of spaces. If $G$ is grouplike then you can say "every" instead of "some" (as long as you remember to specify also that $X$ is not empty!).
Call these the "weak principal homogeneous spaces". Note that if $Y\to X$ is a weak equivalence of $G$-spaces and $X$ is of this kind, then $Y$ is as well.
I think the question might be something like this:
We would like to
(1) specify which weak principal homogeneous spaces will be allowed as fibers
(2) specify what we mean by "fibration" (locally trivial bundle? Serre fibration? quasifibration? ...)
and then consider as "principal $G$-fibrations" those maps $E\to B$ with fiber-preserving right $G$-action such the map is as in (2) and the fibers are as in (1), and then be able to say:
Homotopy classes of maps $B\to BG$ correspond bijectively with equivalence classes of principal $G$-fibrations on $B$. This requires that we also
(3) specify what we mean by an equivalence between two such principal fibrations on the same base.
When $G$ is a topological group then the standard thing is to say (1) the fibers should be isomorphic to $G$ as $G$-spaces (2) locally trivial fiber bundle (and local triviality respecting the $G$-action follows), (3) isomorphism.
If you want to stick with "isomorphism" for (3) in the more general case, then:
Since there is only one homotopy class $\star\to BG$, for (1) you are committed to choosing a single $G$-space $X$ and allowing as fibers only things isomorphic to $X$.
And by considering bundles over disks it seems that you are also committed to choosing "locally trivial bundle" in (2) (there are locally trivializations respecting the $G$-action).
And it seems that this $X$ had better be such that its automorphism group, say $\Gamma$, is also weakly equivalent to $G$, or more precisely such that when considered as a left $\Gamma$-space $X$ is a weak principal homogeneous space.
It will not work to choose $G$ itself as $X$ unless the group of invertible elements of $G$ is equivalent to $G$.
You are in luck if, for example, $G$ is a topological monoid that happens to admit a weak equivalence $G\to \Gamma$ to a topological group such that $\Gamma$ is algebraically generated by the image. But this is rare.
There is probably more then one right answer. Maybe you can allow all weak principal homogeneous spaces as fibers and use Serre fibrations (or maybe quasifibrations) and let equivalence between two such things over the base $B$ mean a map (respecting the map to $B$ and the $G$-action) that is a weak equivalence of total spaces, or equivalently of fibers. Does anybody know?
Note: There is always a group $\Gamma$ related to $G$ indirectly by weak equivalences $G\leftarrow ? \to \Gamma$, so that $BG\simeq B\Gamma$ represents principle $\Gamma$-bundles, but I don't think that's the kind of answer that's wanted.