From the literature on Banach lattice algebras (and that on ordered Banach algebras) there does not appear to be a consensus on a definition.
What is agreed is that one should be a Banach lattice, be an associative algebra with a sub-multiplicative norm and that the product of positive elements should be positive. To standardise terminology we propose that a Banach lattice algebra simply is at the same time a Banach lattice, an algebra with the sub-multiplicative norm, and with the product of positive elements being positive.
Question: For $X$ being a Banach lattice algebra endowed with an ordering $\leq$, let $X_+:=\{x\in X:x>0\}$ be the positive cone. This cone would generally be open or closed? Please provide proof or hints to prove it with references.
Actually, I think the set you defined as $X_+$ is neither closed nor open. Let me change the notation slightly: $X$ is the normed space, $P=\{x\in X: x\geq 0\}$ its positive cone and $P_+=P\setminus \{0\}$ the set of non zero positive elements. Your question regards the topological properties of $P_+$ (although there is also a chance that you mistook $P$ for $P_+$).
First of all, in every Banach lattice, the set $P$ is a closed proper subset. Since normed spaces are connected, it can not be open.
Concerning $P_+$, to see that it is not closed, just pick a nonzero $x_0\in P$ and consider the sequence $(\tfrac{1}{n}x_0)_{n\in \mathbb{N}}$ which is a sequence in $P_+$ that converges to $0\notin P_+$.
Now to show that $P_+$ is not open, you need a little extra work.
1. If $P$ is not a half line, we just need to find a nonzero boundary point of $P$. Lets pick an $x_0\in P_+$ and a $y_0\in X\setminus P$ such that $0$ does not belong in the line segment that connects $x_0$ to $y_0$. It should be easy to find such a pair $(x_0, y_0)$: Just pick a $y_0\in X\setminus P$, set $L=\{\lambda y_0: \lambda \in \mathbb{R}\}$ to be the line passing through $0$ and $y_0$ and pick an $x_0\in P\setminus L$. This set is nonempty since $P$ is not a half line.
Now define $f: [0,1]\rightarrow X$, with $f(\lambda)=\lambda x_0+(1-\lambda)y_0$, for $\lambda\in [0,1]$. Clearly, $f(0)=y_0\in X\setminus P$ and $f(1)=x_0\in P$. Set $\lambda_0=\inf\{\lambda\in [0,1]: f(\lambda)\in P\}$. Since $P$ is closed, $f(\lambda_0) \in P$. Additionally, $f(\lambda) \in X\setminus P$, for every $\lambda<\lambda_0$, since $P$ is convex.
This imples that $f(\lambda_0)\in \partial P$, $f(\lambda)\neq 0 \forall \lambda$, so $f(\lambda_0)$ can't be an interior point of $P_+$.
2. If $P$ is a half line, then your original space can't be a lattice space: Since $P$ is a half line, $P=\{\lambda x_0: \lambda\geq 0\}$, for some nonzero $x_0\in X$. Then for $x, y\in X$,
$$x\leq y \iff y-x=\lambda x_0, $$ for some $\lambda \geq 0$. Now pick $y_0, z_0\in X$ such that the set $\{x_0, y_0, z_0\}$ is linearly independent. Claim: The set of common upper bounds of $y_0$ and $z_0$ is empty: Suppose that $w\in X $ is a common upper bound. Then $$w-z_0=\lambda x_0 \ \ \text{ and }\ \ w-y_0=\lambda' x_0,$$ for some $\lambda, \lambda'\geq 0$. Then $y_0-z_0+(\lambda'-\lambda)x_0=0$, which contradicts the linear independency of these three vectors.