Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in the Brauer group.
To get a division algebra that is not isomorphic to its opposite algebra we can use an element of order 3 in the Brauer group. One method for constructing those is to start with a (cyclic) Galois extension of number fields $E/F$ such that $[E:F]=3$. Let $\sigma\in Gal(E/F)$ be the generator. Let $\gamma\in F$ be an element that cannot be written in the form $\gamma=N(x)$, where $N:E\rightarrow F, x\mapsto x\sigma(x)\sigma^2(x)$ is the relative norm map. Consider the set of matrices
$$
\mathcal{A}(E,F,\sigma,\gamma)=\left\{
\left(\begin{array}{rrr}
x_0&\sigma(x_2)&\sigma^2(x_1)\\
\gamma x_1&\sigma(x_0)&\sigma^2(x_2)\\
\gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0)
\end{array}\right)\mid x_0,x_1,x_2\in E\right\}.
$$
A theorem of A. Albert tells us that this forms a division algebra with center $F$, and its order in $Br(F)$ is 3, so it will not be isomorphic to its opposite algebra. The theory is described for example in ch. 8 of Jacobson's Basic Algebra II. The buzzword 'cyclic division algebra' should give you some hits.
For a concrete example consider the following. Let $F=\mathbf{Q}(\sqrt{-3})$ and let
$E=F(\zeta_9)$, with $\zeta_9=e^{2\pi i/9}$. Then $E/F$ is a cubic extension of cyclotomic fields, $\sigma:\zeta_9\mapsto\zeta_9^4$. I claim that the element $2$ does not belong the image of the norm map. This follows from the fact 2 is totally inert in the extension tower $E/F/\mathbf{Q}$. Basically because $GF(2^6)$ is the smallest finite field of characteristic 2 that contains a primitive ninth root of unity. Now, if $2=N(x)$ for some $x\in E$, then 2 must appear as a factor (with a positive coefficient) in the fractional ideal generated by $x$. But the norm map then multiplies that coefficient by 3, and as there were no other primes above 2, we cannot cancel that. Sorry, if this is too sketchy.
Anyway (see Jacobson again), the product of the $\gamma$ elements modulo $N(E^*)$ is the operation in the Brauer group $Br(E/F)\le Br(F).$ Therefore the opposite algebra should correspond to the choice $\gamma=1/2$, (or to the choice $\gamma=4$, as $2\cdot4=N(2)$). So
$$
\mathcal{A}(E,F,\sigma,2)^{opp}\cong
\mathcal{A}(E,F,\sigma,1/2).
$$
and the choices $\gamma=2$ and $\gamma=1/2$ yield non-isomorphic division algebras, as their ratio $=4$ is not in the image of the norm map.
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Edit (added more details here, because another answer links to this answer): In general the cyclic division algebra construction works much the same for any cyclic extension $E/F$. When $[E:F]=n$ we get a set of $n\times n$ matrices with entries in $E$. The number $\gamma$ appears in the lower diagonal part of the matrix. The condition for this to be a division algebra is that $\gamma^k$ should not be a norm for any integer $k, 0<k<n$. Obviously it suffices to check this for (maximal) proper divisors of $n$. In particular, if $n$ is a prime, then it suffices to check that $\gamma$ itself is not a norm.
Edit^2: Matt E's answer here linear algebra over a division ring vs. over a field gives a simpler cyclic division algebra of $3\times3$ matrices with entries in the real subfield of the seventh cyclotomic field.
Let $A$ be a finite commutative ring (not assumed to contain an identity). Suppose that $a \in A$ is not a zero-divisor. Then multiplication by $a$ induces an injection from $A$ to itself, which is necessarily a bijection, since $A$ is finite. Thus multiplication by $a$ is a permutation of the finite set $A$, and hence multiplication by some power of $a$ (which by associativity is the same as some power of multiplication by $a$) is the identity permutation of $A$. That is, some power of $a$ acts as the identity under multiplication, which is to say, it is a (and hence the) multiplicative identity of $A$.
In short, if a finite commutative ring $A$ contains a non-zero divisor, then it necessarily contains an identity, and every non-zero divisor in $A$ is a unit.
Best Answer
In order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other (standard) ring axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray}\rm (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x \,+\, \color{#C00}{x\!+\!y} \,+\, y\\ \rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x\, +\, \color{#0A0}{y\!+\!x}\, +\, y\end{eqnarray}\bigg\rbrace\:\Rightarrow\: \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
Said more structurally, recall that a SemiRing is that generalization of a Ring whose additive structure is relaxed from a commutative Group to merely a SemiGroup, i.e. here the only hypothesis on addition is that it be associative (so in SemiRings, unlike Rings, addition need not be commutative, nor need every element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may be stated as follows: a semiring with $\,1\,$ and cancellative addition has commutative addition. Such semirings are simply subsemirings of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative semigroup embeds canonically into a commutative group, its group of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$ i.e. the additive version of the fraction field construction).
Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals; distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$; $\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings (e.g. formal power series); formal languages with union, concat; etc. For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory. pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong, Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz. Mathematics and its Applications, 336. Kluwer Academic Publishers Group, Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel. North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7 Zbl review, AMS review