We need the sequence $(|c_{k+1}-c_k|)_k$ to have $0$ Cesaro sum. More specifically,
Let $S_n=\sum_{k=1}^nc_kz^k$ where $c_k\in\mathbb C$ and
$$\frac1n\sum_{k=1}^n|c_{k+1}-c_k|\to0 \tag{$\star$},$$
then $\frac1nS_n\to0$ for all $|z|\leq1$ with $z\neq1$.
First, some remarks:
$(c_k)$ being Cesaro summable is not enough, consider $(c_k)=(1,2,1,2,...)$ and $z=-1$.
$(c_k)$ is convergent$\;\Rightarrow(\star)$, but the reverse is false, consider $c_k=\sqrt k$.
$(\star) \nRightarrow (c_k)$ is Cesaro summable or bounded
, consider $c_k=\sqrt k$.
$|c_{k+1}-c_k|\to0\Rightarrow(\star)$, but reverse is false, consider $c_k=1$ if $k$ is a square and $c_k=2$ otherwise.
$(c_k)$ is Cesaro summable and $\big|\frac{c_{k+1}}{c_k}\big|\to1\Rightarrow(\star)$, but reverse is false, consider the same example in the above line.
Proof sketch:
Let $S_n(z)=\sum_{k=1}^nc_kz^k$ and it follows
\begin{align*}
S_n(z)&=c_nz^n+c_{n-1}z^{n-1}+\cdots+c_1z\\[.6em]
&=z^n(c_n-c_{n-1})+\big(z^n+z^{n-1}\big)(c_{n-1}-c_{n-2})+\cdots+\big(z^n+z^{n-1}+\cdots+z^2\big)(c_2-c_1)\\
&\ \ \ \ +\big(z^n+\cdots+z\big)c_1\\[.5em]
&=z^n(c_n-c_{n-1})+\frac{1-z^2}{1-z}z^{n-1}(c_{n-1}-c_{n-2})+\cdots+\frac{1-z^{n-1}}{1-z}z^2(c_2-c_1)+\frac{1-z^n}{1-z}zc_1.
\end{align*}
Since $|z|\leq1$ we have $|1-z^k|\leq2$ for all $k$ and therefore
$$\frac1n|S_n(z)|\leq\frac2{1-z}\left(\frac{c_1}n+\frac1n\sum_{k=1}^{n-1}|c_{k+1}-c_k|\right).$$
As $n\to\infty$, the right hand side tends to $0$ due to $z\neq1$ and the condition $(|c_{k+1}-c_k|)$ has $0$ Cesaro sum.
For a sanity check, note that without $(c_k)$ we have $\frac1n\sum z^k\to0$ since the 'directions' of all $z^k$ cancel each other out. With the weight $(c_n)$, the directions may not cancel out since some directions can have larger weights than others. Hence, to ensure $\frac1n\sum c_kz^k\to0$ we need $(c_k)$ to be sufficiently 'uniform', in the sense that the weights for all directions are roughly the same in average. This loosely justifies the condition $|c_{k+1}-c_k|$ having $0$ Cesaro sum.
Also, boundedness of $c_k$ seems rather irrelevant here, since $c_k$ being bounded doesn't tell us anything about whether $c_k$ are biased towards some directions. (Unless you want to control the magnitude of the weights)
