As can be easily shown, we know that $Ae = e$, in which $e$ denotes the all-ones vector of appropriate size. Thus, $1 \in \sigma(A)$.
Now, let $(\lambda,v)$ be an eigenpair of $A$, in which $A$ is stochastic. Without loss of generality, we may assume that
$$ \vert \vert v \vert \vert_\infty := \max_{1 \le k \le n} \{ \vert v_k \vert \} = 1.$$
Thus, there is a positive integer $i$, $1 \le i \le n$, such that $\vert v_i \vert = 1$. Since $Av = \lambda v$, by the mechanics of matrix multiplication and the above, we have
$$\vert \lambda \vert =
\vert \lambda \vert\cdot 1 =
\vert \lambda \vert \cdot \vert v_i \vert =
\vert \lambda v_i \vert = \left\vert \sum_{j=1}^n a_{ij} v_j \right \vert
\le \sum_{j=1}^n a_{ij} \vert v_j \vert
\le \sum_{j=1}^n a_{ij} = 1,
$$
i.e., $\vert \lambda \vert \le 1$.