Localization is a technique which allows one to concentrate attention to what is happening near a prime, for example. When you localize at a prime, you have simplified abruptly the behavior of your ring outside that prime but you have more or less kept everything inside it intact.
For lots of questions, this significantly simplifies things.
Indeed, there are very general procedures, in lots of contexts, which go by the name of localization, and their purpose is usually the same: if you are lucky, the problems you are interested in can be solved locally and then the "local solutions" can be glued together to obtain a solution to your original problem. Moreover, an immense deal of effort has been done in order to extent the meaning of "local" so as to be able to apply this strategy in more contexts: I have always loved the way the proofs of some huge theorems of algebraic geometry consist more or less in setting up an elaborate technology in order to be able to say the magical "It is enough to prove this locally", and then, thanks to the fact that we worked so much in that technology, immediately conclude the proof with a "where it is obvious" :)
Of course, all sort of bad things can happen. For example, sometimes the "local solutions" cannot be glued together into a "global solution", &c. (Incidentally, when this happens, so that you can do something locally but not glue the result, you end up with a cohomology theory which, more or less, is the art of dealing with that problem)
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.
Best Answer
Let $R$ be a ring without unit. Suppose
We claim that $R$ is a ring with unit.
Let $b \in R$, $b \ne 0$. Then there is a unique $e_b$ such that $b = be_b = e_bb$.
We must show that $e_a = e_b$ for all nonzero $a,b$. Then this will be the unit in $R$.
Let $a,b \in R$, both nonzero. There is $c$ so that $a = cb = bc$. So $$ e_b a = e_b b c = b c = a,\qquad ae_b= c b e_b = c b = a. $$ Thus, $e_b$ satisfies the defining property of $e_a$. By the uniqueness, $e_a = e_b$.