The suspensions $\Sigma({\mathbb R}P^3)$ and $\Sigma({\mathbb R}P^2\vee S^3)$ are not homotopy equivalent and can be distinguished by the Pontryagin square $H^2(X;{\mathbb Z}_2)\to H^4(X;{\mathbb Z}_4)$. One reference for this is the book "Combinatorial Homotopy and 4-Dimensional Complexes" by Hans Baues, where this fact is stated in an example on page 20 and proved later in the book using the theory developed there. Baues also cites Hilton's book "Homotopy Theory and Duality" for another proof. I would guess that this fact was known to J.H.C.Whitehead when he wrote his two papers "On simply connected 4-dimensional polyhedra" (1949) and "A certain exact sequence" (1950) which develop a general classification of simply-connected 4-complexes. It may also have been known earlier to Pontryagin. I wonder if there are other modern expositions of this material.
If one suspends again to $\Sigma^2({\mathbb R}P^3)$ and $\Sigma^2({\mathbb R}P^2\vee S^3)$ then these spaces become homotopy equivalent. The attaching map of the top cell of $\Sigma^n{\mathbb R}P^3$ gives an element of $\pi_{n+2}(\Sigma^n{\mathbb R}P^2)$ and for $n\geq 2$ this group can be computed to be ${\mathbb Z}_2$ generated by the suspended Hopf map $S^{n+2}\to S^{n+1}\subset \Sigma^n{\mathbb R}P^2$ by looking at the long exact sequence of (stable) homotopy groups for the cofiber sequence $S^{n+1}\to S^{n+1}\to \Sigma^n{\mathbb R}P^2$ where the first map has degree 2. Thus there are two possible attaching maps for the top cell of $\Sigma^n{\mathbb R}P^3$, up to homotopy, and they are distinguished by whether the Steenrod square $Sq^2$ is trivial on $H^{n+1}(\Sigma^n{\mathbb R}P^3;{\mathbb Z}_2)$ or not, using the fact that $Sq^2$ is nontrivial on the mapping cone of the suspended Hopf map. For ${\mathbb R}P^3$ the action of $Sq^2$ is trivial for dimension reasons, hence the action is also trivial for $\Sigma^n{\mathbb R}P^3$ so the top cell must be attached trivially.
Some further interesting information: The Hopf map $S^3\to S^2$ has infinite order, but when when we attach a 3-cell to $S^2$ by a map of degree 2 to form $\Sigma{\mathbb R}P^2$, the Hopf map still generates $\pi_3(\Sigma{\mathbb R}P^2)$ but now has order 4, rather than 2 as one might guess. More generally, if we attach the 3-cell by a map of degree $d$ then in the resulting 3-complex the Hopf map has order $2d$ if $d$ is even, but order $d$ if $d$ is odd. Going back to the case at hand, we have $\pi_3(\Sigma{\mathbb R}P^2)={\mathbb Z}_4$ and the attaching map of the 4-cell of $\Sigma{\mathbb R}P^3$ is the element of order 2 in this ${\mathbb Z}_4$. (It can't be a generator since $Sq^2$ acts trivially.)
Let $\Lambda$ be a module over a principal ideal domain $R$. The Künneth Theorem states that
$$H^n(X\times Y; \Lambda) \cong \bigoplus_{p+q=n}H^p(X; \Lambda)\otimes H^q(Y; \Lambda) \oplus \bigoplus_{p+q=n+1}\operatorname{Tor}(H^p(X; \Lambda), H^q(Y; \Lambda)).$$
If $\Lambda = R$ and $R$ is a field, then $\operatorname{Tor}(H^p(X; R), H^q(Y; R)) = 0$ as $H^p(X; R)$ and $H^q(Y; R)$ are free $R$-modules (i.e. vector spaces over $R$). For $\Lambda = R = \mathbb{Z}$, there may be some contribution from the $\operatorname{Tor}$ terms.
In this case, $X = Y = S^2$ and
$$H^p(S^2; \mathbb{Z}) = \begin{cases}
\mathbb{Z} & p = 0, 2\\
0 & \text{otherwise}.
\end{cases}$$
As $H^p(S^2; \mathbb{Z})$ is a free $\mathbb{Z}$-module for every $p$, we see that all the $\operatorname{Tor}$ terms vanish and therefore
$$H^n(S^2\times S^2; \mathbb{Z}) = \begin{cases}
\mathbb{Z} & n = 0, 4\\
\mathbb{Z}\oplus\mathbb{Z} & n = 2\\
0 & \text{otherwise}.
\end{cases}$$
As for the wedge sum, what you stated is not correct, it only holds for reduced cohomology. That is, $\widetilde{H}^n(X\vee Y; \mathbb{Z}) \cong \widetilde{H}^n(X;\mathbb{Z})\oplus\widetilde{H}^n(Y;\mathbb{Z})$. So your calculation of the cohomology groups is correct, except in degree zero:
$$H^n(S^2\vee S^2\vee S^4; \mathbb{Z}) \cong \begin{cases}
\mathbb{Z} & n = 0, 4\\
\mathbb{Z}\oplus\mathbb{Z} & n = 2\\
0 & \text{otherwise}.
\end{cases}$$
As $H^n(S^2\times S^2; \mathbb{Z}) \cong H^n(S^2\vee S^2\vee S^4; \mathbb{Z})$ for every $n$, your argument for showing that $S^2\times S^2$ and $S^2\vee S^2\vee S^4$ are not homotopy equivalent doesn't work.
So far we've only considered the cohomology groups of the two spaces, but one can also consider their cohomology rings.
Let $X$ be a topological space. For any principal ideal domain $R$, cup product endows $H^*(X; R) := \bigoplus_{n\geq 0}H^n(X; R)$ with the structure of a graded ring. If $f : X \to Y$ is a continuous map, then $f^* : H^*(Y; R) \to H^*(X; R)$ is a ring homomorphism, in particular, if $f = \operatorname{id}_X$, then $f^* = \operatorname{id}$. It follows that if $X$ and $Y$ are homotopy equivalent, then $X$ and $Y$ have isomorphic cohomology rings, not just cohomology groups.
If $\alpha, \beta$ denote the two generators of $H^2(S^2\times S^2; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$, then $\alpha\cup\beta \neq 0$. On the other hand, if $\gamma, \delta$ denote the two generators of $H^2(S^2\vee S^2\vee S^4; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$, then $\gamma\cup\delta = 0$. Therefore $S^2\times S^2$ and $S^2\vee S^2\vee S^4$ do not have isomorphic cohomology rings, so they are not homotopy equivalent.
Best Answer
Their homology groups with $\mathbb{Z}$-coefficients are different. Same for cohomology. (As long as $n>1$ — if $n=1$, the spaces are homeomorphic.)