Is there a way to reconstruct a rectangular prism with this information?

Sure. In a system of equations, you need at least as many equations as you have variables in order to determine values for all variables. Since you have 3 variables and 3 equations, you should be able to get values for (a,b,c).

So, the new question is, how do I?

The first place I would start is ( E / 4 ) = A + B + C and ( S / 2) = AB + AC + AB and then plug in the known values for E and S.

Starting at where @jerv left off, what would happen if we keep subtracting the first (of those two) equations from the second until we get zero on one side?

(S / 2) – N(E / 4) = AB + BC + CA – NA – NB – NC
(S / 2) – N(E / 4) = A(B-N) + B(C-N) + C(A-N)

If N = 2S/E, then

0 = A(B – 2S/E) + B(C – 2S/E) + C(A – 2S/E)

I don’t know if this helps.

hmm, what if you separate ‘a’ and constants to one side of the equations?

v/a = bc

e/4-a = b + c

substitute in:
s = 2a(b+c) + 2bc

s = 2a(e/4-a) + 2(v/a)

The 3 sides are the roots of
x^3 – k1 x^2 + k2x – k3 where
k1 = a+b+c
k2 = ab + ac + bc
k3 = abc

@LostInParadise Actually, a quick test shows that you need to change that to
– x^3 + k1 x^2 – k2x + k3.

It works, but I don’t know why.

If you take (x – a)(x-b)(x-c)=0 then the trinomial you get has coefficients that satisfy the conditions I gave. You negated all the coefficients that I gave. The solution is the same when set to 0.

@LostInParadise I think you’re on the right track—i.e., it reduces to a cubic & your formulation has pleasing symmetry.