The definition of Cesaro summability you give is for that of the series $\ S_n=\sum_\limits{i=1}^na_i\ .$ Kemeny and Snell's theorem $\ 5.1.4\ $ does not say that the series $\ \sum_\limits{i=1}^nP^i\ $ is Cesaro summable. Part (a) says
The $\color{red}{\it{sequence}}$ $\ P^n\ $ is Cesaro-summable to $\ A\ .$
According to the standard definition of Cesaro summability of sequences, as also given by Kemeny and Snell themselves (on p.18 of their first edition of 1960) this means that
$$
A=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^nP^i\ .
$$
As you've already discovered, for your matrix $\ P\ $ the Cesaro sum of the sequence $\ \big\{P^n\big\}\ $ is
$$
A=\pmatrix{\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}}\ ,
$$
but the series $\ \sum_\limits{i=1}^nP^i\ $ is not Cesaro summable.
Part (b) of Kemeny and Snell's theorem $\ 5.1.4\ $ says
The $\color{red}{\it{series}}$ $\ I+\sum_\limits{i=1}^n\big(P^i-A\big)\ $ is Cesaro-summable to $\ Z\ .$
where $\ Z\ $ has been defined earlier (on p.100 of the first edition) as $\ (I+A-P)^{-1}\ .$ This means that
$$
Z=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^n\left(I+\sum_\limits{i=1}^j\big(P^i-A\big)\right)\ ,\tag{1}\label{e1}
$$
but says nothing whatever about the Cesaro summability of the series $\ \sum_\limits{i=1}^nP^i\ .$
For your matrix $\ P\ ,$ we have
\begin{align}
Z&=\pmatrix{\frac{3}{2}&-\frac{1}{2}\\-\frac{1}{2}&\frac{3}{2}}^{-1}\\
&=\pmatrix{\frac{3}{4}&\frac{1}{4}\\\frac{1}{4}&\frac{3}{4}}\ ,
\end{align}
and
$$
\sum_{i=1}^{j}\big(P^i-A\big)=\cases{\pmatrix{-\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}}&if $\ j\ $ is odd\\
\pmatrix{0&0\\0&0}&if $\ j\ $ is even.}
$$
You might like to check that the identity (\ref{e1}) does indeed hold for your matrix $\ P\ .$
Note also that the definition of ergodicity for Markov chains given by Kemeny and Snell is much less common nowadays than it was in the $1960$s. The definition used by most modern texts requires a chain to be aperiodic to be classified as ergodic.