Note that
$$ \mathrm{tr}_B\left(|0\rangle \langle1|\otimes|0\rangle \langle1|\right) =|0\rangle \langle1| \mathrm{tr} (|0\rangle \langle1|) = |0\rangle \langle1| \, \langle1|0\rangle = |0\rangle \langle1| \cdot 0 = 0 $$
so the two terms marked red below disappear when you take the partial trace
$$ W_{A}=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|+\color{red}{\frac{s}{2}|0\rangle \langle1|+\frac{s}{2}|1\rangle \langle0|}. $$
Once these terms are gone, the rest of the calculations yields
$$ \begin{align} W_{A}&=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|\\ &=\frac{1}{2}|0\rangle \langle0|+\frac{1}{2}|1\rangle \langle1|+\frac{s}{2}|0\rangle \langle1|\\ &=\frac{1}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \end{align} $$$$ \begin{align} W_{A}&=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|\\ &=\frac{1}{2}|0\rangle \langle0|+\frac{1}{2}|1\rangle \langle1|\\ &=\frac{1}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \end{align} $$
as expected.
There is another way of computing the partial trace from the full density matrix that works particularly well in this case
$$ W_A = \mathrm{tr}_B\begin{pmatrix} (1+s)/4 & 0 & 0 & s/2 \\ 0 & (1-s)/4 & 0 & 0 \\ 0 & 0 & (1-s)/4 & 0 \\ s/2 & 0 & 0 & (1+s)/4 \\ \end{pmatrix} = \begin{pmatrix} \mathrm{tr}\begin{pmatrix} (1+s)/4 & 0 \\ 0 & (1-s)/4 \end{pmatrix} & \mathrm{tr}\begin{pmatrix} 0 & s/2 \\ 0 & 0 \end{pmatrix} \\ \mathrm{tr}\begin{pmatrix} 0 & 0 \\ s/2 & 0 \end{pmatrix} & \mathrm{tr}\begin{pmatrix} (1-s)/4 & 0 \\ 0 & (1+s)/4 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}. $$