It's not a scalar product at all, because it's not positive definite. Physicists call it a scalar product since it has all algebraic rules in common with a scalar product; it's only lacking the positive definiteness. It's called a symmetric bilinear form. Given an arbitrary basis ##\vec{e}_{\mu}##, a symmetric bilinear form is a function ##V \times V \rightarrow \mathbb{R}##, ##\vec{x},\vec{y} \mapsto \vec{x} \cdot \vec{y}##, where ##V## is a real vector space, is completely characterized by a matrix
$$g_{\mu \nu} = \vec{e}_{\mu} \cdot \vec{e}_{\nu}.$$
By assumption it obeys ##\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}## (that's, why it's called a symmetric bilinear form). One can show there exists always a basis ##\vec{e}_1,\ldots,\vec{e}_d## (##d##: dimension of the vector space) such that ##(g_{\mu \nu})=\mathrm{diag}(1,1,\ldots,-1,-1,\ldots,0,0,\ldots)## (with of course ##d## entries in total).
A bilinear form is called non-degenerate, if there are no zeroes in the above formula, and a symmetric non-degenerate bilinear form is called a "fundamental form" of the vector space. It's characterized by the number of ones in the above diagonal matrix (then the remaining entries are of course just ##-1##'s). A Lorentz bilinear form (or Lorentz product) has (in your convention) 3 ones and 1 ##-1## in the diagonal matrix, and that's precisely what you need for special relativity: It's a 4D real vector space with a fundamental bilinear form with the signature (3,1) (giving the number of ##+1##'s and ##-1##'s in the diagonal matrix).
Why does one need now precisely this form? That's, because of Einstein's postulate about the invariance of the speed of light in a vacuum under "boosts", where boosts are such transformations of the spatial coordinates and the time which describe the transformation from one inertial reference frame to another, without rotating the spatial axis, i.e., the one is simply moving with a constant velocity against another.
To figure out the transformations, describing such a boost was done by Einstein in 1905 in a quite cumbersome way (but you should try to understand it, because it's very physical compared to the one I summarize here). This derivation is much simplifight by the ingeneous insight by Minkowski of 1908, that one can describe special relativistic spacetime as a vector space (when fixing just an arbitrary spacetime point as an "origin") with exactly a Lorentzian fundamental bilinear form! Choosing an inertial frame then means that it's exactly one of the preferred bases which make the scalar product look simple, i.e.,
$$\vec{e}_{\mu} \cdot \vec{e}_{\nu} = \eta_{\mu \nu} = \mathrm{diag}(1,1,1,-1).$$
The vector components with respect to such a basis are then
$$(\boldsymbol{x},c t),$$
where ##\boldsymbol{x}## is the position vector of some point and ##t## the time an observer measures with an ideal clock who is at rest relative to this coordinate system.
Now think about a flashlight, located in the origin of the coordinate system, which is switched on and off again after a short while. Then a shperical light-pulse will spread out, which is located at the spherical shell given by the equation ##\boldsymbol{x}^2=c^2 t^2## (with the usual 3D Euclidean scalar product for the spatial 3D vector) since after a time ##t## the pulse, traveling with the speed of light in a vacuum ##c##, has reached a distance of ##c t## from the origin. This we can of course write in terms of the Minkowski product as
$$\vec{x} \cdot \vec{x}=\boldsymbol{x}^2-c^2 t^2=0.$$
You can now characterize the Lorentz boosts as such linear transformations (or the corresponding change of the four basis vectors) that don't rotate the spatial axes against each other, for which the Minkowski product reads precisely the same. This socalled "covariant" notation makes life in relativity much easier than any other notation.
For more on this (but with the other convention of the sign of the Minkowski product), see
http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf