I will first describe the naive correspondence that is assumed in usual literature and then I will say why it's wrong (addressing your last question about hidden assumptions) :)
The postulate of relativity would be completely empty if the inertial frames weren't somehow specified. So here there is already hidden an implicit assumption that we are talking only about rotations and translations (which imply that the universe is isotropic and homogenous), boosts and combinations of these. From classical physics we know there are two possible groups that could accomodate these symmetries: the Gallilean group and the Poincaré group (there is a catch here I mentioned; I'll describe it at the end of the post). Constancy of speed of light then implies that the group of automorphisms must be the Poincaré group and consequently, the geometry must be Minkowskian.
[Sidenote: how to obtain geometry from a group? You look at its biggest normal subgroup and factor by it; what you're left with is a homogeneous space that is acted upon by the original group. Examples: $E(2)$ (symmetries of the Euclidean plane) has the group of (improper) rotations $O(2)$ as the normal subgroup and $E(2) / O(2)$ gives ${\mathbb R}^2$. Similarly $O(1,3) \ltimes {\mathbb R}^4 / O(1,3)$ gives us Minkowski space.]
The converse direction is trivial because it's easy to check that the Minkowski space satisfies both of Einstein postulates.
Now to address the catch: there are actually not two but eight kinematical groups that describe isotropic and uniform universes and are also consistent with quantum mechanics. They have classified in the Bacry, Lévy-Leblond. The relations among them is described in the Dyson's Missed opportunities (p. 9). E.g., there is a group that has absolute space (instead of absolute time that we have in classical physics) but this is ruled out by the postulate of constant speed of light. In fact, only two groups remain after Einstein's postulate have been taken into account: besides the Poincaré group, we have the group of symmetries of the de Sitter space (and in terms of the above geometric program it is $O(1,4) / O(1,3)$).
Actually, one could also drop the above mentioned restriction to groups that make sense in quantum mechanics and then we could also have an anti de Sitter space ($O(2,3) / O(1,3)$). In fact, this shouldn't be surprising as general relativity is a natural generalization of the special relativity so that the Einstein's postulates are actually weak enough that they describe maximally symmetric Lorentzian manifolds (which probably wasn't what Einstein intented originally).
A synchronisation procedure of ideal clocks at mutual rest must be transitive, symmetric, reflexive and it must remain valid in time once one has adjusted the clocks to impose it. There is no evident a priori reasons why Einstein's procedure should satisfy these constraints. The fact that it instead happens is the physical content of both postulates you quoted.
Actually there is a third physical constraint: The value of the velocity of light must be constantly $c$ when measured along a closed path. This measurement does not need a synchronisation procedure, since just one clock is exploited.
A natural issue show up at this juncture: whether there are synchronisation procedures different from Einstein's one which however fulfill all requirements.
The answer is positive (without imposing other constraints like isotropy and homogeneity) and they give rise to other formulations of special relativity, which are physically equivalent to Einstein's one. (Geometrically speaking, it turns out that the geometry of the rest spaces is not induced by the metric of the spacetime by means of the standard procedure of induction of a metric on a submanifold.)
There are well-known physical situations regarding clocks with non-inertial motion (at rest with respect to each other), where Einstein's procedure cannot be used and other synchronisation procedures must be adopted. The most relevant is the one regarding a rotating platform. If I remember well, the first correct analysis of the problem was proposed by Born.
Best Answer
Light doesn't travel at $c+V$ (where $V$ is the speed of the source), it travels at $c$.
What's the difference? It means that if you're flying towards someone at a speed $V$ and you shine a light at them, you measure the light to travel away from you at a speed $c$, but the other person measures it to fly past them also at a speed $c$ (i.e. not $V+c$).
In the case of sound, the source and observer may disagree on the relative speed. The source will measure the sound to propagate at a speed $U-V$, whereas the observer will measure it to propagate at $U$.