Are all non-associative (not necessarily associative) finite division rings finite fields?
Consider the algebra over $\mathbb{F}_3$ with basis (as a vector space over $\mathbb{F}_3$) the set $\{1,x,x^2\}$ and multiplication given by: \begin{eqnarray*} x(x^2)&=&x+2,\\ (x^2)x&=&1+x+x^2,\\ (x^2)(x^2)&=&x. \end{eqnarray*}
By construction it is finite, has a two-sided identity $1$ and multiplication distributes over addition. The first two equations demonstrate that it is non-associative. Also left or right multiplication by any fixed non-zero element is bijective (see proof below).
I fixed the first equation and did a computer search through the $676$ possibilities for the other two. Of these $14$ came out as having the left and right cancellation property. One of these was of course $\mathbb{F}_{27}$. The other $13$ are non-associative, and of them the above algebra seemed like the nicest.
Proof of left and right cancellation:
It suffices to prove that left multiplication by any non-zero element is injective, as then it must also be surjective and the algebra will not contain non-zero zero-divisors. Thus right multiplication by any non-zero element would also be injective, hence bijective.
Both $x^3-x^2-x-1$ and $x^3-x-2$ are irreducible over $\mathbb{F}_3$ as they have no roots in $\mathbb{F}_3$.
Left multiplication by a non-zero $\mathbb{F}_3$-linear combination $\alpha(x)$ of $1$ and $x$ is the same map as left multiplication by $\alpha(x)$ in $\mathbb{F}_3[x]/(x^3-x-2)\cong\mathbb{F}_{27}$ - hence bijective.
Similarly left multiplication by a non-zero $\mathbb{F}_3$-linear combination $\alpha(y)$ of $1$ and $y=x^2$ is the same map as left multiplication by $\alpha(y)$ in $\mathbb{F}_3[y]/(y^3-y^2-y-1)\cong\mathbb{F}_{27}$ - hence bijective.
Thus without loss of generality, if there is a non-zero left zero-divisor, there will be one of the form $\lambda+x\pm x^2$, for some $\lambda\in \mathbb{F}_3$. Thus it suffices to check that the matrices representing left multiplication by $x\pm x^2$ have no eigenvalues in $\mathbb{F}_3$. The characteristic polynomials of these matrices are:
$$ \left| \begin{array}{ccc} t&2 &1 \\ 2&t+2&1\\ 2&1&t \end{array}\right| = t^3-t^2-t-1 ,\qquad \left| \begin{array}{ccc} t&1 &1 \\ 2&t+1&0\\ 1&0&t \end{array}\right| = t^3+t^2+2 .$$
Neither of these cubics have roots in $\mathbb{F}_3$.
We provide a family of examples of finite division algebras, which additionally have a right identity. This does not answer the revised version of the question which asks for a finite division algebra (other than a finite field) which has a two-sided identity.
On any finite field $\mathbb{F}_q$ with $q=p^r$, and $p$ prime, $r>1$, we can define $a\star b= ab^p$. This is non-commutative ($a\star b\neq b\star a \iff a^{-1}b\notin \mathbb{F}_p$), but has the two-sided cancellation property: $$a\star b=0\implies a=0\,\, {\rm or}\,\, b=0,$$ and has a right identity.