How do I prove that $S^1\vee S^2\vee S^3$ and $S^1\times S^2$ are not homotopy equivalent using homology and cohomology ring respectively?

If you're allowed to do cohomology, note that $H^*(S^1 \vee S^2 \vee S^3)$ is isomorphic to the direct sum $H^*(S^1)\oplus H^*(S^2) \oplus H^*(S^3)$, so cup product of the generators of $H^1$ and $H^2$ is trivial. Whereas cup product of the generators of $H^1$ and $H^2$ is evidently nontrivial for $S^1 \times S^2$, by Kunneth formula if you wish.

Doing this with homology is a bit tricky: note that universal cover of $S^1 \vee S^2 \vee S^3$ is the topological space which looks like $\Bbb R$ with a copy of $S^2 \vee S^3$ attached to each integer point in $\Bbb R$. This is homotopy equivalent to an infinite wedge of $S^2 \vee S^3$'s, hence the homology group $H_3$ is nontrivial - in fact isomorphic to $\bigoplus \Bbb Z$

On the other universal cover of $S^1 \times S^2$ is $\Bbb R \times S^2$, which is homotopy equivalent to $S^2$. That has trivial homology on dimension $3$, so is not homotopy equivalent to the universal cover of $S^1 \vee S^2 \vee S^3$. Thus, $S^1 \times S^2$ is not homotopy equivalent to $S^1 \vee S^2 \vee S^3$ either.

it is interesting to note after suspending once they are homotopy equivalent, in effect suspending kills the ring structure DS

If they were homotopy equivalent, there would be a map $S^1 \vee S^2 \vee S^3 \to S^1 \times S^2$ inducing an isomorphism on homology. The hard part is clearly $S^3$, so let's focus on that.

Because $S^3$ is simply connected, any map $S^3 \to S^1 \times S^2$ lifts to the universal cover $\Bbb R \times S^2$, which deformation retracts onto $\{0\} \times S^2$. Therefore any map $S^3 \to S^1 \times S^2$ is homotopic to one which factors through $\{1\} \times S^2$. In particular, it induces the trivial map on $H_3$, as $H_3(S^2;A) = 0$ for any coefficients $A$.

Because a map from the wedge sum is just a map from each factor sending the basepoint of each to the same place, this means that any map $S^1 \vee S^2 \vee S^3 \to S^1 \times S^2$ induces the trivial map on $H_3$ with any coefficients. So it cannot be a homology isomorphism, and definitely not a homotopy equivalence.