See the title. This is true if the sequence is nonnegative; some Tauberian theorems which I was able to find give some more general sufficient conditions. I would like to know if this is true for arbitrary bounded sequences.
Recall that for a sequence $(a_n)$ with natural indices $n$, the Cesàro means are $\frac1{N}\sum\limits_{n=1}^N a_n$, and the Abel means are $(1-r)\sum\limits_n r^n a_n$.
Yes. For the following modes of convergence you can prove $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)$.
$$\begin{eqnarray} a_n &\to& a \qquad (1)\\ \sigma_n:={1\over n}\sum_{j=1}^{n} a_j &\to& a \qquad (2)\\ (1-r)\sum_{n=1}^\infty a_n r^n &\to& a\qquad (3)\\ (1-r)\sum_{n=1}^\infty \sigma_n r^n &\to& a\qquad (4) \end{eqnarray} $$
If $a_n$ is a bounded sequence then $(2)\Longleftrightarrow (3)$. One direction is $(2)\Rightarrow (3)$ and the other follows from Littlewood's Tauberian theorem ($(4)\Rightarrow (2)$) since $\sigma_{n+1}-\sigma_n$ is $O(1/n)$.
Reference. For further information and two proofs of the "Abel to Cesàro" theorem, see Chapter 1, sections 7, 11, and 12 of Tauberian Theory: A Century of Development by Jacob Korevaar.