This has been bugging me for a while any help would be appreciated.
The second bullet point from this nLab page says:
Let $A$ be a non-associative unital algebra with finite dimension, then it's possible to find a case (over $R$) where $A$ has no zero divisors, but there exists a non-zero element in A that has no inverse (i.e. nonzero $x$, where $xa = ax = 1$).
However this MSE post says:
Let $A$ be a non-associative (although power-associative) unital algebra with finite dimension. Then if $A$ has no divisors implies every nonzero element in $A$ has an inverse (particularly looking at the proof by Robert Lewis).
Is there a contradiction somewhere, or am I overlooking a use of associativity or technicality?
In finite-dimensional unital algebras, being free of zero divisors is equivalent to being a division algebra.
It's not possible if the algebra under consideration is power-associative or flexible, for instance. Because it's a division algebra it must have even dimension bounded by 8. You ask for the right-inverse and left-inverse to be the same, that is the reason you can construct such algebra, we only need to build a division algebra with at least one element whose right and left inverses do not coincide.
In this article it's shown a general method to obtain division algebras using the Cayley-Dickson construction.
For an example, follow this answer and consider $\mathrm{Cay}(\mathbb{H},i)$ given by the vector space of pairs of Quaternions (the usual, generated by $1,i,j,k$) and with multiplication given by:
$(u,v)\cdot(u',v') = (u\cdot u'+ i(\bar{v'}v), v'u + v\bar{u'})$
It's a particular case of the construction above so the article deals with the affirmation "it is a division algebra".
Now take $v=(0,1+i$). The right-inverse for $v$ must satisfy: $(0,1+i)\cdot(x,y) =(i(\bar{y}(1+i)),(1+i)\bar{x})=(1,0)$ implying $x=0$ and $y=\frac{-1+i}{2}$. It can not be a left-inverse because $(0,\frac{-1+i}{2})\cdot(0,1+i)=(-1,0)$.