There is the following result on the completion of a normed algebra:
Suppose $X$ is a normed algebra. Then there is a Banach algebra $Y$ and a map $T$ from $X$ onto a dense subspace $M \subset Y$ such that the map $T$ is an isometric isomorphism for which $T(x_1 \cdot x_2) = T(x_1) \cdot T(x_2)$ whenever $x_1, x_2 \in X$.
The following should prove the above mentioned result:
Let $X$ be a normed algebra. Then, in particular, $X$ is a normed space, and hence there is an isometric isomorphism $T$ from $X$ into a Banach space $Y$ such that $T(X)$ is dense. Since $T(Y) \subset Y$ is dense, there are, for any $x, y \in Y$, sequences $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}} \subset X$ such that $x = \lim_{n \in \mathbb{N}} Tx_n$ and $y = \lim_{n \in \mathbb{N}} Ty_n$.
Define now $x \cdot y = \lim_{n \in \mathbb{N}} T(x_n \cdot y_n)$, which is well-defined since $(T(x_n \cdot y_n))_{n \in \mathbb{N}}$ is Cauchy in $Y$.
The above stated proof is rather straightforward. However, I am wondering if there are other ways to complete a normed algebra.
Thanks in advance; any comments are welcome.
In your proof, you still have to show that the choice of sequences $(x_n)$ and $(y_n)$ converging to $x$ and $y$ do not interfere in the definition of $xy$. You also need to verify that this multiplication satisfies all the necessary properties (associativity, scalars moving inside products, etc...), although this is easy.
As a comment, there are other ways of constructing $Y$. You can take $\ell^\infty(\mathbb{N},X)$ to be the algebra of bounded sequences in $X$ with supremum norm, and $c_0(\mathbb{N},X)$ the ideal of those sequences converging to $0$. $\ell^\infty(\mathbb{N},X)$ is not necessarily complete, but the quotient $Y=\ell^\infty(\mathbb{N},X)/c_0(\mathbb{N},X)$ is (the proof is not hard, but a little cumbersome), and it is a Banach algebra.
The map $X\to Y$, sending $x\in X$ to the class of the constant sequence $(x,x,\ldots)$ is an isometric homomorphism, so the closure of its image has the desired properties.