EDIT (came back to make this rigorous as I realized that the reasoning is a bit more complicated).
Consider the following
Lemma. Let $G$ be a (compact) group with unitary representation $\Pi$ on finite-dimensional Hilbert space $\mathscr{H}$ and let $\Pi(G)^\perp$ denote all the linear operators on $\mathscr{H}$ which commutes with $\Pi(G)$. If $\Pi(G)\cap \Pi(G)^\perp = \mathbb{C}$ is trivial (i.e., contains only scalar identities), then $\mathscr{H} \cong \mathscr{A}\otimes \mathscr{B}$ where $(\mathscr{A},\Pi_\mathscr{A})$ is an irreducible representation of $G$, i.e., $\Pi(g) \cong \Pi_\mathscr{A}(g)\otimes I_\mathscr{B}$ for any $g\in G$, and any $\Pi(G)^\perp \cong I_\mathscr{A} \otimes \mathscr{L}(\mathscr{B})$ where $\mathscr{L}(\mathscr{B})$ denote all linear operators on $\mathscr{B}$.
Using the lemma, the statement isn't too hard to show, so let's prove the lemma. Since $\Pi$ is a unitary representation, it's completely reducible so that
$$
\mathscr{H} =\bigoplus_{V_k} V_k^{\oplus n_k}
$$
where the summation is over non-isomorphic irreps $V_k$ of $G$ and $n_k$ denotes the number of copies. Note that $V_k^{\oplus n_k} \cong V_k \otimes \mathbb{C}^{n_k}$.
Therefore, we must show that the direct sum is over a single irrep $V_k$. Indeed, suppose that the direct sum is over multiple irreps. Then note that the operator $\bigoplus_{V_k} c_k I_{V_k \otimes \mathbb{C}^{n_k}}$ where $c_k\in \mathbb{C}$ are arbitrarily chosen, commutes with $\Pi(G)$ and thus we reach a contradiction. Hence, we see that the summation can only be over a single irrep, i.e.,
$$
\mathscr{H} \cong \mathscr{A}\otimes \mathscr{B}
$$
where $\mathscr{A}$ is an irrep of $G$.
Since $\mathscr{A}$ is an irrep of $G$, we see that any linear operator on $\mathscr{A}$ is a linear combination of operators in $\Pi(G)$, (see for example here), and since $\Pi(G)^\perp\cap \Pi(G) =\mathbb{C}$, we see that $\Pi(G)^\perp \cong I_{\mathscr{A}} \otimes \mathscr{L}(\mathscr{B})$
