What you're missing is that $[\mathbb{S},\mathbb{S}]=\mathbb{Z}$. Let now $R$ be the infinite matrix of integers representing $\rho$. Note that since $\rho$ takes value in $\bigoplus_{I_1}\mathbb{Z}\subseteq \prod_{I_1}\mathbb{Z}$, all of its columns have only finitely many non-zero values. So, for every column of $R$ we can construct a map $\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$ sending $1\in\pi_0(\mathbb{S})$ to the column in $\bigoplus_{I_1}\mathbb{Z}=\pi_0\bigvee_{I_1}\mathbb{S}$ (for example by composing $\mathbb{S}\to \bigvee_{I_1'}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$, where $I'_1$ is the finite subset of $I_1$ where the column is non-zero and using that finite wedges are categorical products and so $[\mathbb{S},\bigvee_{I_1'}\mathbb{S}]=\prod_{I_1'}[\mathbb{S},\mathbb{S}]=\prod_{I_1'}\mathbb{Z}$).
Finally, using that the wedge is the categorical coproduct we can put it all together in a map $$\bigvee_{I_2}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$$ as required.
In fact what I'm doing is essentially proving that the map $$\pi_0:\left[\bigvee_{I_2}\mathbb{S},\bigvee_{I_1}\mathbb{S}\right]\to \mathrm{Hom}\left(\bigoplus_{I_2}\mathbb{Z},\bigoplus_{I_1}\mathbb{Z}\right)$$ is an isomorphism (injectivity is quite easy to show).