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I think this isn't true, but I am not sure how to prove it ?

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    I would look at the homology. – 2017-02-10
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    Yeah never mind I got it. The fundmental group of the torus is $Z \times Z$ while the fundmental group of $S^1 \vee S^1$ is just free groups on two generators. – 2017-02-10

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For the sake of completeness: Suppose that a retraction $r \colon S^1 \times S^1 \to S^1 \vee S^1$ exists, so that $r \circ i = \operatorname{Id}_{S^1 \vee S^1}$, where $i$ is an inclusion of $S^1 \vee S^1$ into $S^1 \times S^1$. Then $r_* \circ i_* = \operatorname{Id}_{\pi_1(S^1 \vee S^1)}$. It follows that $r_*$ is surjective and $i_*$ is injective.

Let $a$ and $b$ be generators of $\pi_1(S^1 \vee S^1)$ and note that $x = aba^{-1}b^{-1}$ is not the identity, as $\pi_1(S^1 \vee S^1)$ is free. We see that $$i_*(x) = i_*(aba^{-1}b^{-1}) = i_*(a)i_*(b)i_*(a^{-1})i_*(b^{-1}) = i_*(a)i_*(b)i_*(a)^{-1}i_*(b)^{-1}$$ which, by commutativity of $\pi_1(S^1 \times S^1)$ is $i_*(a)i_*(a)^{-1}i_*(b)^{-1}i_*(b)^{-1} = 0$. This contradicts the fact that $i_*$ is injective.