I think the following can be turned into a proof, but I haven't checked the details.

By a result of Milnor, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $F(S^0 \vee S^0)$, the free group functor on the pointed set $S^0 \vee S^0$. By a version of the Hilton-Milnor theorem, the latter coincides up to homotopy with $F(S^0) \ast F(S^0)$, the latter being the free product with amalgamation of simplicial free groups. But $F(S^0)$ coincides with $\Bbb Z$ up to homotopy. Hence, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $\Bbb Z \ast \Bbb Z$ (as a space). In particular, $\Omega (S^1 \vee S^1)$ is homotopically discrete, which gives the result you are after (in conjunction with the van Kampen theorem).

I think the following can be turned into a proof, but I haven't checked the details.

By a result of Milnor, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $F(S^0 \vee S^0)$, the free group functor on the pointed set $S^0 \vee S^0$. By a version of the Hilton-Milnor theorem, the latter coincides up to homotopy with $F(S^0) \ast F(S^0)$, the latter being the free product with amalgamation of simplicial free groups. But $F(S^0)$ coincides with $\Bbb Z$ up to homotopy. Hence, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $\Bbb Z \ast \Bbb Z$ (as a space). In particular, $\Omega (S^1 \vee S^1)$ is homotopically discrete, which gives the result you are after (in conjunction with the van Kampen theorem).