If f is the identity map on C, set g = f to demonstrate chain equivalence. Thus C is homotopy equivalent to itself.
By definition, C is homotopy equivalent to D iff D is homotopy equivalent to C.
The composition of two chain equivalences gives another chain equivalence. Verify this in steps. If f and g are the successive chain equivalences, let f′ and g′ be their inverses. With gg′ homotopic to I, fgg′ is homotopic to fI. Put f′ on the right and fgg′f′ is homotopic to ff′, which is homotopic to I. The same holds in the reverse direction, hence fg is a chain equivalence.
Being homotopy equivalent is, well, an equivalence relation, as its name suggests. Thus each chain complex represents a specific homotopy type.
It's homology groups may not be the same as those of C. In fact, each Ci×S could drop to 0, giving a 0 homology.
Assume C and E are homotopy equivalent, as demonstrated by the chain maps f and g. Tensor C, E, f, and g with S. Apply f×S and g×S to get an endomorphism on C×S. Let d be the homotopy that equates fg with the identity map on C. Tensor d with S to build the new homotopy that equates (f×S)(g×S) with the identity map on C×S. In other words, C×S and E×S are homotopy equivalent, and have the same homology.