Let $M$ be a connected, compact, and orientable 3-manifold ($H^3(M)\cong\mathbb{Z}$), and let $G$ be a simple Lie group satisfying $\pi_1(G)=\pi_2(G)=0$. Let
$\pi_M(G)$ denote the set of homotopy classes of maps from $M\to G$
$\pi_3(G)$ denote the set of homotopy classes of maps from $\mathbb{S}^3\to G$
Suppose I have a map $f:M\to \mathbb{S}^3$ which induces an isomorphism on $H^3$, and for which $f_*:\pi_3(G)\to \pi_M(G)$ is surjective. $$f_*[g]:=[g\circ f]$$
Is this enough information to infer that $f_*$ is an isomorphism?
If $G$ is a simple Lie group, then $\pi_3(G) = \Bbb Z$ (see here). So the only way this homomorphism could fail to be an isomorphism is if $\pi_M(G)$ were finite. Now $G$ is homotopy equivalent to a space $G'$ with a single 0-cell, a single 3-cell, 4-cells whose boundary maps are the constant maps to the basepoint, and cells of higher dimension; this follows from theorem 4C.1 in Hatcher.
From this, cellular approximation shows that every map $M \to G'$ is homotopic to one lying entirely in the copy of $S^3$; Hopf's theorem says that these are classified by their degree; and therefore even after wedging on the extra copies of $S^4$ they are still not homotopic. (This is because the inclusion $S^3 \hookrightarrow G'$ induces an isomorphism on $H^3$.)
So there are infinitely many homotopy classes of maps $M \to G$ as desired. Your homomorphism is an isomorphism.