Sorry to be late to the party, but here's an example of 2 compact simply connected manifolds which have the same homotopy groups, same homology groups, same cohomology ring, and yet are not homotopy equivalent. The examples are motivated by Grigory M's examples:$S^2\times S^2$ and $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$. His examples are both $S^2$ bundles over $S^2$.
If we extend this further, it turns out there are precisely two $S^3$ bundles over $S^2$. Of course, one is the product $S^3\times S^2$, while another doesn't have a more common name, so I'll just denote it $S^3\hat{\times} S^2$.
Both of these spaces are diffeomorphic to quotients of free linear $S^1$ actions on $S^3\times S^3$. Letting $X$ denote either bundle, we have a long exact sequence of homotopy groups $$...\pi_k(S^1)\rightarrow \pi_k(S^3\times S^3)\rightarrow \pi_k(X)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$$
which can be used to show that $\pi_k(X) \cong \pi_k(S^3\times S^3)$ for $k\geq 2$ and $\pi_1(X) = \{e\}$ and $\pi_2(X) \cong \mathbb{Z}$.
The Hurewicz theorem together with Universal coefficients theorem implies $H^1(X) = 0$ and $H^2(X) \cong \mathbb{Z}$. Poincare duality then forces the rest of the cohomology rings to agree.
Finally, to see $S^3\times S^2$ and $S^3\hat{\times} S^2$ are different, one computes the Stiefel-Whitney classes of their tangent bundles. It turns out $w_2(S^3\times S^2) = 0$ while $w_2(S^3\hat{\times}S^2)\neq 0$. (And all other Stiefel-Whitney classes are $0$ for both spaces). Since the Stiefel-Whitney classes can be defined in terms of Steenrod powers, they are homotopy invariants, so $S^2\times S^3$ and $S^3\hat{\times}S^2$ are not homotopy equivalent.
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
The space $S^1\vee S^1\vee S^2$ is the wedge sum of two circles and one sphere. In particular this space is also known as mouse space.
It is quite easy to see that the fundamental group of $S^1\vee S^1\vee S^2$ is $\mathbb{Z}*\mathbb{Z}\cong\langle a,b| \emptyset \rangle$ ( you can use the Seifert-VanKampen theorem to see it ); whereas the fundamental group of the torus is $\mathbb{Z}\times\mathbb{Z}=\langle a,b | [a,b]=1\rangle$, which is not isomorphic to the previous.
If there was a homotopic equivalence between them; their fundamental groups needs to be isomorphic; but this is not true. So there is not any homotopic equivalence.