To show that the space is complete, one needs to demonstrate that all Cauchy sequences with respect to the norm converge. If $(a_n)_{n\geq 0}\subseteq \ell^1(I;\mathbb{R})$ is such a Cauchy sequence then by definition, for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\|a_n-a_m\|<\epsilon$ whenever $n,m\geq N$.
Assume first of all that $I$ is a countable indexing set. Then, for each fixed $i\in I$ and all $n,m\geq N$, one has
$$|a_n(i)-a_m(i)|\leq \sum_{i\in I} |a_n(i)-a_m(i)| \leq \|a_n-a_m\| <\epsilon.$$
This means that for each fixed $i\in I$, the sequence $(a_n(i))_{n\geq 0}\subseteq \mathbb{R}$ is a Cauchy sequence with respect to $|\cdot|$. Since $(\mathbb{R},|\cdot|)$ is complete, for each fixed $i\in I$ this sequence converges, that is, $\lim_{n\to\infty} a_n(i)=a(i)$ for some $a(i)\in\mathbb{R}$.
Define the sequence $a=(a(i))_{i\in I}$, it is a candidate for the limit of $(a_n)_{n\geq 0}$. We wish to show that $a\in\ell^1(I;\mathbb{R})$. For this, we observe firstly that for a fixed $n\geq N$ we have $a-a_n\in \ell^1(I;\mathbb{R})$. Notice that since $I$ is countable, we can write it as $I=\{i_j\}_{j\in\mathbb{N}}$ and so
\begin{align*}
\sum_{i\in I} |a(i)-a_n(i)|&= \sum_{j=1}^\infty |a(i_j)-a_n(i_j)|\\
&=\lim_{J\to\infty}\sum_{j=1}^J |a(i_j)-a_n(i_j)|\\
&=\lim_{J\to\infty}\sum_{j=1}^J \lim_{m\to\infty}|a_m(i_j)-a_n(i_j)|\\
&=\lim_{J\to\infty}\lim_{m\to\infty}\sum_{j=1}^J|a_m(i_j)-a_n(i_j)|\\
&\leq \lim_{J\to\infty}\lim_{m\to\infty} \|a_m-a_n\|\\
&<\lim_{J\to\infty} \epsilon = \epsilon.
\end{align*}
In the above, the third equality is by definition of $a_m(i_j)\to a(i_j)$ for each $i_j$, the fourth equality is interchanging a limit and finite sum, then the first inequality is obtained by bounding the finite sum by the full sum over $I$ which is just the norm, and the final inequality is obtained from the fact that this norm is bounded by $\epsilon$ when $n,m\geq N$ ($n$ was chosen larger than $N$ and $m$ gets larger than $N$ in the limit to infinity). By definition, we then have that $a-a_n\in\ell^1(I;\mathbb{R})$. But then notice that by the triangle inequality and the fact that $a_n\in\ell^1(I;\mathbb{R})$ we have
$$\sum_{i\in I}|a(i)|\leq \sum_{i\in I} |a(i)-a_n(i)|+\sum_{i\in I} |a_n(i)|<\infty,$$
and thus $a\in\ell^1(I;\mathbb{R})$. The fact that $a_n\to a$ then follows by observing that the set of equations above showed that for arbitrary $\epsilon>0$, we have found for all $n\geq N$ that $\|a-a_n\|<\epsilon$. Thus the Cauchy sequence converges and the space is complete.
Now we turn to the case where $I$ is uncountable. One can show then that if a sequence $a$ is in $\ell^1(I,\mathbb{R})$, then $a(i)\neq 0$ for at most countably many $i\in I$. If you are stuck with why this is the case see the following answer. Now, if $(a_n)_{n\geq 0}$ is a Cauchy sequence in $\ell^1(I;\mathbb{R})$, there exists a countable subset $J\subseteq I$ such that for all $n\geq 0$,
$$a_n(i)=0 \ \ \forall\ i\in I\setminus J.$$
To see that this is the case, we have by the claim above that for each $a_n$, there exists a countable set $J_n\subseteq I$ for which $a_n(i)=0$ whenever $i\in I\setminus J_n$. Let $J=\bigcup_{n=0}^\infty J_n$ so that $J$ is a countable set being a countable union of countable sets. Then, For any $n\geq 0$, we have $a_n(i)=0$ whenever $i\in I\setminus J$ since $I\setminus J\subseteq I\setminus J_n$.
This means that for each of the sequences $a_n$ in the Cauchy sequence, we have
$$\|a_n\|=\sum_{i\in I}|a_n(i)|=\sum_{j\in J}|a_n(j)|,$$
with this latter sum being a countable sum. Using the exact same working as in the countable index case above for this sequence then gives the claim for the case when $I$ is uncountable.
