Is spacetime defined mathematically without using $c$ speed? Is there a mathematical definition of spacetime that does not use $c$ speed as a conversion factor or involve the spacetime interval?  If not why?
 A: 
Is there a mathematical definition of spacetime that does not use c speed as a conversion factor or involve the spacetime interval? 

Yes, absolutely! The quantity $c$ really is just a units-conversion factor. If we use a coordinate system $t,x,y,z$ with $t$ expressed in units of time and $x,y,z$ expressed in units of length, then we need to use $c$ in some equations in order to equalize the units, like in this equation for the proper time $\tau$:
$$
   d\tau^2 = dt^2 - \frac{dx^2+dy^2+dz^2}{c^2}.
\tag{1}
$$
(Here, $d\tau$ is the proper-time increment along one piece of the object's history, and $dt,dx,dy,dz$ are the corresponding coordinate-increments.) But we can also use a coordinate system $t,x,y,z$ (I'm recycling the letters) with all four coordinates expressed in the same units. Then equation (1) becomes
$$
   d\tau^2 = dt^2 - (dx^2+dy^2+dz^2).
\tag{2}
$$
In fact, this is the way physicists usually do things in general relativity (as Ben Crowell's comment said). We can either express $\tau,t,x,y,z$ all in units of time, or we could express them all in units of length.
The justification for this is essentially the same as the justification for expressing both vertical distances and horizontal distances in the same units. We could express vertical distances in meters and express horizontal distances in feet, but then we would need to include an awkward units-conversion factor in the equation for the length-increment $d\ell$ in 3-d space, like this:
$$
 d\ell^2 = \frac{dx^2+dy^2}{a^2}+dz^2
\tag{3}
$$
where $a$ is the conversion factor that relates horizontal- and vertical-distance units. Thanks to rotational symmetry, though, it is more natural to use the same units for both types of distance — especially because most distances are neither vertical nor horizontal, but somewhere in between. So we might as well use the same units for both and just write the distance equation like this:
$$
 d\ell^2 = dx^2+dy^2+dz^2.
\tag{4}
$$
The difference between equations (1) and (2) is the same as the difference between equations (3) and (4). In the case of equations (1) and (2),  Lorentz symmetry is the reason why (2) is more natural than (1), juse like rotational symmetry is the reason why (4) is more natural than (3). Lorentz symmetry mixes the coordinate $t$ with the coordinates $x,y,z$, just like rotational symmetry mixes the coordinates $x,y$ with the coordinate $z$.
