It's just not true that having isomorphic Lie algebras implies a bijection between the irreducibles (presumably you mean a bijection compatible with the isomorphism between the Lie algebras). For example when $G = SU(2), H = SO(3)$ only half of the irreducibles of $SU(2)$ come from irreducibles of $SO(3)$.
Continuing, the natural double cover $SU(2) \to SO(3)$ identifies the algebra $C(SO(3))$ of continuous functions on $SO(3)$ with a subalgebra of $C(SU(2))$ (exactly the subalgebra of "even functions" with respect to $-1 \in SU(2)$) and in the Peter-Weyl decomposition this subalgebra corresponds to the irreducibles of $SU(2)$ which descend to irreducibles of $SO(3)$ ("integer spin"), while there's a whole other subspace of odd functions corresponding to the irreducibles which don't ("half-integer spin").
Generally, if $G$ and $H$ are compact connected Lie groups related by a covering map $p : G \to H$ then they fit into a short exact sequence
$$1 \to Z \to G \to H \to 1$$
where $Z = \text{ker}(p)$ is a finite central subgroup of $G$, which when $G$ is simply connected can be identified with $\pi_1(H)$. When $G = SU(2), H = SO(3)$ we have $Z = \{ \pm 1 \}$. The pullback $p^{\ast} : \text{Rep}(H) \to \text{Rep}(G)$ then identifies the representations of $H$ with the representations of $G$ on which $Z$ acts trivially; meanwhile there must be other representations of $G$ on which $Z$ doesn't act trivially (by Peter-Weyl). This pullback is compatible with the pullback $p^{\ast} : C(H) \to C(G)$ on continuous functions, which identifies $C(H)$ with the subalgebra of $C(G)$ on which $Z$ acts trivially (the generalized "even subalgebra").