As per Mark Grant's suggestion, here are some additional details for my comment.

We may write $\Bbb P^1$ with its circle action as the adjunction space $$* \cup_{S^1 \times 0} S^1 \times [0,1] \cup_{S^1 \times 1} *,$$

with the circle acting in the obvious way on $S^1 \times [0,1]$. Passing to the Borel construction, we find that $$\Bbb P^1 \times_{S^1} ES^1 = BS^1 \cup_{ES^1 \times 0} ES^1 \times [0,1] \cup_{ES^1 \times 1} BS^1.$$

The picture is that we are connecting two copies of $BS^1$ by a contractible bit.

To make this precise, collapse the closed contractible subspace $ES^1 \times 1/2$ to a point. As this is contractible, the projection map is a homotopy equivalence; thus $\Bbb P_{hS^1}$ is homotopy equivalent to the wedge of two mapping cones $$\text{Cone}(ES^1 \to BS^1) \vee \text{Cone}(ES^1 \to BS^1).$$ Because $ES^1$ is contractible, each of these cones are homotopy equivalent to $BS^1$ itself.

Thus we see that $$\Bbb{CP}^\infty \vee \Bbb{CP}^\infty \simeq \Bbb P^1_{hS^1};$$ further the projection map $\Bbb P^1_{hS^1} \to \Bbb{CP}^\infty$ sends the two wedge summands identically onto $\Bbb{CP}^\infty$.

Thus $$H^*_{S^1}(\Bbb P^1;\Bbb Z) \cong \Bbb Z[x,y]/(xy),$$ with action of $u \in H^2_{S^1}(pt)$ given by $u \cdot x = x^2, u \cdot y = y^2.$ This is isomorphic as a graded module to $$H^*(\Bbb P^1;\Bbb Z) \otimes H^*_{S^1}(pt;\Bbb Z),$$ but not as an algebra: the equivariant cohomology has no nilpotent elements, whereas the tensor-product algebra does, given by the generator of $H^2(\Bbb P^1) \otimes H^0_{S^1}(pt)$.