Euler characteristic expression in terms the number of fixed points of an $\mathbb{S}^1$ action

This is true for any smooth circle action on a closed manifold. There is no need to have an almost complex structure.

In fact, more generally we have the following.

Suppose $S^1$ acts smoothly on a closed manifold $M$. Then the fixed point set $M^{S^1}$ has the property that $\chi(M) = \chi(M^{S^1})$.

(So, in particular, the result holds even if the fixed point set is not isolated).

Proof: By averaging an arbitrary Riemannian metric on $M$, we may assume the action is isometric. The fixed point set of an isometric action is always a totally geodesic submanifold, but it may have several components. (It can only have finitely many components since it is compact). In particular, asking about $\chi(M^{S^1})$ makes sense.

We also point out that in this situation, the number of isotropy groups is finite. (This is true, up to conjugacy, for any compact Lie group action on any closed manifold, as a consequence of the slice theorem).

Let $N$ be a component of $M^{S^1}$ and let $\nu N$ be an embedding of the normal bundle of $N$ into some $\epsilon$-neighborhood of $N$. By shrinking $\epsilon$, we may assume that $\nu N_1 \cap \nu N_2 = \emptyset$ for disjoint components $N_1,N_2\subseteq M^{S^1}$. I will use the notation $\nu M^{S^1}$ to denote the the union of the $\nu N_i$

Note that the $S^1$ action preseves $\nu M^{S^1}$ since is is characterized as the set of points a distance $< \epsilon$ away from $M^{S^1}$ and the action is isometric.

It follows that the $S^1$ action also preserves $M\setminus \nu M^{S^1}$. Since we've removed the points with isotropy group $S^1$, and all other closed subgroups of $S^1$ are finite (and there are only finitely many of them) there is a neighborhood $U$ of the identity $1\in S^1$ with the property that any $p\in U$ acts with no fixed points. In particular, the action field ( $\frac{d}{dt}|_{t=0} e^{it} \cdot m$, for $m\in M$) has no zeros. By Poincare-Hopf, $M\setminus \nu M^{S^1}$ has zero Euler characteristic.

In a similar fashion, the $S^1$ action preserves the boundary $\partial \nu M^{S^1}$ since $\partial \nu M^{S^1}$ consists of all points in $M$ a distance $\epsilon$ from $N$. Repeating the argument in the previous paragraph, we deduce $\partial \nu N$ also has zero Euler characteristic.

Now we are basically done. Write $M = (M\setminus \nu M^{S^1}) \cup \nu M^{S^1}$. Using the fact that each $\nu N_i$ deformation retracts to $N_i$ (so, in particular, $\chi(\nu M^{S^1}) = \chi(M^{S^1})$, we compute \begin{align*} \chi(M) &= \chi(M\setminus \nu M^{S^1}) + \chi(\nu M^{S^1}) - \chi(\partial \nu M^{S^1})\\&= 0 + \chi(\nu M^{S^1}) + 0 \\ &= \chi(M^{S^1}). \end{align*}