Ignoring monodromy gives the wrong Serre E2 page

Reflection sends the positive loop $t\mapsto [t]$ to the negative loop $t\mapsto [-t]$. Hence [F1] gives $r_{\ast }=-1:H_1(S^1;\mathbb{Z})⟶H_1(S^1;\mathbb{Z})\mathrm{.}$ By [F2], this is exactly the degree-one monodromy of the mapping-torus fibration. On $H_0(S^1;\mathbb{Z})=\mathbb{Z}$ the monodromy is the identity, because the fiber is connected.

In fiber degree one, the cellular local chain complex of the one-vertex, one-edge base circle is $0⟶\mathbb{Z}\stackrel{1-(-1)=2}{\to }\mathbb{Z}⟶0.$ Changing the edge convention multiplies this boundary by $-1$ and changes neither group. Thus [F3] gives $E_{0,1}^2=\mathrm{coker}(2)=\mathbb{Z}\mathrm{/}2,E_{1,1}^2=\ker (2)=0.$ If one falsely makes the coefficient system constant, the boundary becomes $1-1=0$, so the same two entries become $\mathbb{Z}$ and $\mathbb{Z}$.

Apply [F2] in degree one. On fiber $H_1$, $1-r_{\ast }=2$; on fiber $H_0$, $1-r_{\ast }=0$. Hence $0⟶\mathbb{Z}\mathrm{/}2⟶H_1(K;\mathbb{Z})⟶\mathbb{Z}⟶0.$ This sequence splits explicitly: the point $[0]\in S^1$ is fixed by $r$, so $[t]\mapsto [([0],t)]$ is a section of $K\to S^1$ and its first-homology map splits the displayed projection. Therefore $H_1(K;\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}\mathrm{/}2$.

The base has cellular dimension one, so no Serre differential $d_s$ with $s\ge 2$ can leave or enter its columns. The correct $E^2$ table therefore already displays the torsion piece in total degree one. The false constant table has $\mathbb{Z}$ instead at $(0,1)$ and would have two infinite cyclic total-degree-one pieces, so it cannot yield the actual $H_1(K)$. Degree zero, the zero kernel of multiplication by two, identity action on $H_0$, both base cells, both orientations, both row degrees, and the fixed-point section are all explicit above. No AC is used, and there is no converse assertion.