Consider a pair $(X \lor Y, X)$. Let $Z = X - U$. By excision axiom, inclusion $(X \lor Y - Z, X - Z) \subset (X \lor Y, X)$ induces isomorphism in homology. As we have strong deformation retraction of $U$ on $x_0$, and since $(X \lor Y - Z, X - Z) = (U \lor Y, U)$, we get a deformation retraction of a pair $(U \lor Y, U) \to (Y, x_0)$, and the homotopy axiom says that this induces isomorphism in homology. The inverse is given by the map given by inclusion. Overall, we have that inclusion $(Y, x_0) \to (X \lor Y, X)$ induces an isomorphism. This map has an obvious left inverse $(X \lor Y, X) \to (Y, x_0)$ that contracts $X$ to a point. By easy categorical considerations, this map also must induce an isomorphism in homology.
Now consider the long exact sequence for pair $(X \lor Y, X)$:
$ \cdots \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to \tilde{H}_{n-1}(X) \to \cdots$
The map $\tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$ is induced by inclusion $(X, x_0) \to (X \lor Y, x_0)$. There's an obvious left inverse to this inclusion, a map $(X \lor Y, x_0) \to (X, x_0)$ that contracts $Y$ in $X \lor Y$ to a point. As this map has a left inverse, the map induced in homology also has a left inverse, so it's a monomorphism, and we have an exact sequence $0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$
Similarly, consider $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$. As this map is also induced by inclusion $(X \lor Y, x_0) \to (X \lor Y, X)$, and from previous consideration we know that the contraction $(X \lor Y, X) \to (Y, x_0)$ induces an isomorphism, we consider a composition of these two. This map has a right inverse given by inclusion $(Y, x_0) \to (X \lor Y, x_0)$, which induces a left inverse for a map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(Y)$, which necessarily is an epimorphism. Now compose it with the inverse isomorphism $\tilde{H}_n(Y) \to \tilde{H}_n(X \lor Y, X)$ to obtain that $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is also epimorphism, and thus we get an exact sequence $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$. Remember that we composed it with the isomorphism inverse to the one $\tilde{H}_n(X \lor Y, X) \to \tilde{H}_n(Y)$ we used earlier, so here in this exact sequence, the map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is the same as in long exact sequence.
Finally, we connect these two exact sequences to get a short exact sequence:
$0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$
As the first map is induced by inclusion and it has a right inverse we mentioned earlier, we obtain that this s.e.s. split, so $\tilde{H}_n(X \lor Y) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(X \lor Y, X) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(Y)$.
Note that additivity axiom was not used.
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
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.