Let $L$ denote $\mathbb{C}$ with $S^1$ acting by multiplication, and let $\mathbb{C}$ denote $\mathbb{C}$ with trivial $S^1$-action. Then the projective space $P(L\oplus\mathbb{C})$ is homeomorphic to $S^2$, and the natural $S^1$-action is the one that you mentioned. Thus, your problem is a special case of calculating $K_G(PV)$, where $V$ is a complex representation of a compact Lie group $G$. There is an evident map from $R(G)=K_G(\text{point})$ to $K_G(PV)$, and the tautological bundle $T$ also gives an element of $K_G(PV)$, so the polynomial ring $R(G)[T]$ maps to $K_G(PV)$. Put $f(t)=\sum_{k=0}^{\text{dim}(V)}(-1)^k\Lambda^k(V^*)t^k$. The constant bundle with fibre $V$ splits as $T\oplus T^\perp$, and using this one can check that $f(T)=0$ in $K_G(PV)$. With more work it can be shown that $K_G(PV)=R(G)[T]/f(T)$. This is stated as Proposition 3.9 in Segal's "Equivariant K-Theory"; the proof relies on a result
that Segal states as Proposition 3.8, but does not prove; for that, see Proposition 4.9
of Atiyah's "Bott periodicity and the index of elliptic operators". A more direct argument is possible for the case that you mention, but the result above gives the general context.