Is there any difference between the reduced and unreduced suspension? This is related with this arguing here: https://mathoverflow.net/questions/107430/does-the-reduced-mapping-cylinder-have-the-same-homotopy-type-of-unreduced-mappin
The reduced cone has the same homotopy type of the unreduced one, when we are talking about well based spaces. The problem is if the inclusion $X\to CX $, in which $CX$ is the reduced cone, is a unbased cofibration. (I don't think it is in general: only if it is a well pointed space).
Assuming it is well based space, you don't have to worry if it is the reduced suspension or the unreduced one.
Back to the problem:
We are only considering unreduced constructions, as I guess you want.
This is consequence of the excision and the exact sequence.
I guess you are talking about Eilenberg-Steenrod axioms for unreduced homology, since the Eilenberg-Steenrod axioms for the reduced homology assumes the suspension isomorphism as an axiom.
$(\sum X , CX , CX) $ is a excisive triad. By the axiom, you get that
$(CX,X)\to (\sum X, CX)$ induces isomorphisms between the homology groups.
Well, now, using the exact sequence of the pair $(CX, X) $, we can prove that
$ H_q(X, \ast )$ is isomorphic to $H_{q+1}(CX,X) $.
TO prove this last statement, first of all, you have to notice that $ H_q(X)\cong H_q(X, \ast)\oplus H_q(\ast ) $. And, then, notice that that we get a new exact sequence from this congruence. This new exact sequence is
$\cdots \rightarrow H_q(A, \ast )\to H_ q(X, \ast )\to H_q(X,A)\to H_{q-1}(X, \ast)\rightarrow \cdots $.
In our case,
$\cdots \rightarrow H_q(X, \ast )\to H_ q(CX, \ast )\to H_q(CX,X)\to H_{q-1}(X, \ast)\rightarrow \cdots $.
Since $H_ q(CX, \ast )$ is clearly trivial, we conclude that
$H_q(CX,X)\to H_{q-1}(X, \ast) $
is an isomorphism.
So we can conclude that there is an isomorphism $H_q(X, \ast)\to H_ {q+1}(\sum X, CX) $.
Since $ (\sum X, CX)\equiv (\sum X, \ast) $, the required statement was proven.
I'm going to use a different notation mostly because I'm largely copying this out of an old homework of mine.
Let $C_+^n$ and $C_-^n$ be the cones in $\Sigma X$, let $U$ be a neighborhood around the cone point with $\overline{U}\subset Int(C_-^n)$. Then $U$ can be excised, so
$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq \widetilde{H}_q(\Sigma X\setminus U,C_-^n\setminus U).$$
Since $(\Sigma X\setminus U,C_-^n\setminus U)$ deformation retracts to $(C_+^n,X)$ we get
$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq\widetilde{H}_q(C_+^n,X)$$
Since $CX$ is contractible, $C_+^n\simeq C_-^n\simeq \{pt.\}$, and thus $\widetilde{H}_\ast(C_+^n)\simeq\widetilde{H}_\ast(C_-^n)\simeq 0$. Then the long exact sequence
$$ \dots\to \widetilde{H}_q(C_-^n)\to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to\widetilde{H}_{q-1}(C_-^n)\to\dots $$
gives
$$ \dots\to 0 \to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to 0 \to\dots $$
so that $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)$. Similarly, the long exact sequence
$$ \dots\to \widetilde{H}_q(C_+^n)\to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to\widetilde{H}_{q-1}(C_+^n)\to\dots $$
gives
$$ \dots\to 0 \to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to 0 \to\dots $$
so that $ \widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $.
Thus, $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)\simeq\widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $.
Best Answer
All the constructions that you used to define the isomorphism are natural/functorial:
In conclusion, each subsquare in the following diagram is commutative, hence the "outer" rectangle (by inverting the horizontal arrows that go the wrong way, the composite of the whole thing is the suspension isomorphism) is commutative: $$\require{AMScd} \begin{CD} \tilde{H}_i(\Sigma X) @>{\cong}>> \tilde{H}_i(\Sigma X, C_+ X) @<{\cong}<< \tilde{H}_i(C_- X, X) @>{\cong}>> \tilde{H}_{i-1}(X) \\ @VVV @VVV @VVV @VVV \\ \tilde{H}_i(\Sigma Y) @>{\cong}>> \tilde{H}_i(\Sigma Y, C_+ Y) @<{\cong}<< \tilde{H}_i(C_- Y, Y) @>{\cong}>> \tilde{H}_{i-1}(Y) \end{CD}$$
tl;dr The composition of two functors is a functor.