Here is a partial answer to question 1: the necessary and sufficient condition for a (sufficiently reasonable, say a CW-complex) space to be homotopy equivalent to a topological group is that it should have the homotopy type of a loop space, or, in other words, that it should admit a structure of an $A_\infty$-space. The necessity is clear. On the other hand, if $X=\Omega Y$, then Milnor constructs in "The construction of the universal bundles I", section 3, a group $G(Y)$ with the same homotopy type as $X$. The construction is as follows (we assume $Y$ to be a polyhedron): take the subset of the disjoint union of $Y^n,n\geq 1$ formed by all sequences such that any two consecutive elements are in the same simplex and the first and the last elements are the base point, and take the quotient of this subset with respect to the equivalence relation generated by $(x_1,\ldots,x,x,\ldots x_n)\sim (x_1,\ldots,x,\ldots,x_n)$ and $(x_1,\ldots,x,y,x,\ldots x_n)\sim (x_1,\ldots,x\ldots,x_n)$; the product is the concatenation product.
Here are some remarks:
The above is somewhat (but not completely) similar to what happens when one "strictifies" an $A_\infty$ algebra by taking the cobar construction of the bar construction.
An H-space $X$ can have several non-homotopic products. These are one-to-one with $[X\wedge X,X]$, see e.g. Stasheff, H-spaces from a homotopy point of view, p.11, LNM 161 (which also has useful references to earlier work).