Is it true that $K(G \ast H, 1) = K(G,1)\vee K(H,1) $?

Yes, this is true (at least, for nice spaces like CW complexes). As a sketch of a proof, observe that you can construct a universal cover for $X\vee Y$ by gluing together copies of a universal cover of $X$ and a universal cover of $Y$ at preimages of the basepoint. (You can in fact give a combinatorial description of how the copies are glued together which parallels the description of $\pi_1(X)*\pi_1(Y)$ in terms of reduced words.) Since the universal covers of $X$ and $Y$ are contractible, it follows easily that the universal cover of $X\vee Y$ has trivial homology in dimensions greater than $1$, and so it is contractible.

(There are other ways you could do the last step besides using homology; for instance, you could contract the copies of the universal covers of $X$ and $Y$ to a point one by one which does not change the homotopy type since they are contractible; here you use the combinatorics of how they are attached to ensure that no copy can get glued together with itself when you contract other copies.)

Yes, see the question and comments at overflow.

The answer there explains the map $$BG\vee BH \to B(G*H)$$ is a weak equivalence more generally for topological groups, or even topological monoids.