First, $N_1(0) + N_2(0) = 0 + 0 = 0$.
Second, we need to show $\{N_1(t) + N_2(t)\}$ has independent increments, namely, for $0 < t_1 < t_2 < \cdots < t_k$, $N_1(t_1) + N_2(t_1), (N_1(t_2) + N_2(t_2)) - (N_1(t_1) + N_2(t_1)), \ldots, (N_1(t_k) + N_2(t_k)) - (N_1(t_{k - 1}) + N_2(t_{k - 1}))$ are independent, which follows from that $\{N_1(t)\}$ and $\{N_2(t)\}$ are independent and each of $\{N_i(t)\}$ has independent increments.
Third, it needs to be shown $\{N_1(t) + N_2(t)\}$ has stationary increments, namely, for $0 \leq s < t$ and every $n \geq 0$,
$$P((N_1(t) + N_2(t)) - (N_1(s) + N_2(s)) = n) = P(N_1(t - s) + N_2(t - s) = n). $$
Indeed, by the stationary increments property of $N_i(t)$ and the law of total probability, it follows that
\begin{align*}
& P((N_1(t) + N_2(t)) - (N_1(s) + N_2(s)) = n) \\
= & P(N_1(t) - N_1(s) + N_2(t) - N_2(s) = n) \\
= & \sum_{k = 0}^n P(N_1(t) - N_1(s) + N_2(t) - N_2(s) = n|N_2(t) - N_2(s) = k) P(N_2(t) - N_2(s) = k) \\
= & \sum_{k = 0}^n P(N_1(t) - N_1(s) = n - k|N_2(t) - N_2(s) = k) P(N_2(t) - N_2(s) = k) \\
= & \sum_{k = 0}^n P(N_1(t) - N_1(s) = n - k) P(N_2(t) - N_2(s) = k) \text{ by independence of } N_1(t) \text{ and } N_2(t)\\
= & \sum_{k = 0}^n P(N_1(t - s) = n - k)P(N_2(t - s) = k) \\
= & P(N_1(t - s) + N_2(t - s) = n).
\end{align*}
Finally, by the law of total probability, for every $n \in \{0, 1, \ldots\}$,
\begin{align*}
& P(N_1(t) + N_2(t) = n) = \sum_{k = 0}^n P(N_1(t) + N_2(t) = n, N_1(t) = k) \\
= & \sum_{k = 0}^n P\left(N_1(t) + N_2(t) = n\big | N_1(t) = k\right) P(N_1(t) = k) \\
= & \sum_{k = 0}^nP(N_2(t) = n - k|N_1(t) = k)P(N_1(t) = k) \\
= & \sum_{k = 0}^nP(N_2(t) = n - k) P(N_1(t) = k) \quad \text{by independence of } N_1(t) \text{ and } N_2(t) \\
= & \sum_{k = 0}^n e^{-{\lambda_2t}}\frac{(\lambda_2t)^{n - k}}{(n - k)!}\times e^{-\lambda_1t}\frac{(\lambda_1t)^k}{k!} \\
= & \frac{e^{-(\lambda_1 + \lambda_2)}}{n!}\sum_{k = 0}^n \binom{n}{k}(\lambda_2t)^{n - k}(\lambda_1t)^k \\
= & e^{-(\lambda_1 + \lambda_2)t}\frac{((\lambda_1 + \lambda_2)t)^n}{n!}
\end{align*}
Hence $N_1(t) + N_2(t) \sim \text{Poisson}((\lambda_1 + \lambda_2)t))$.