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.
In my opinion Definition 2' is the correct definition (and Definition 2 is not a definition, but rather a characterization), because it follows the general principles how algebraic structures are defined in universal algebra (by operations satisfying equations), and therefore it also directly generalizes to other categories, for example topological spaces. A topological group is not just a topological monoid such that every object has an inverse. We also need that the inverse map is a continuous map (it is an interesting fact that for Lie groups this is automatic). So it is useful when the inverse map belongs to the data.
There are several other reasons why this is important: The subgroup generated by a subset $X$ of a group has underlying set $\{a_1^{\pm 1} \cdots \dotsc \cdots a_n^{\pm 1} : a_i \in X\}$. When you view groups as special monoids, you may forget the $\pm 1$ here.
Different categories should always be considered as disjoint. This helps to organize mathematics a lot. Unfortunately, forgetful functors between these categories are usually ignored, treated as if they were identities, but of course they are not. For example, we have the forgetful functor $U : \mathsf{Grp} \to \mathsf{Mon}$. It turns out that it is fully faithful (in many texts group homomorphisms are defined that way, of course again this is not conceptually correct). It has a left adjoint (Grothendieck construction) as well as a right adjoint (group of units). In particular it preserves all limits and colimits. These properties of $U$ show that often (but not always!) there is no harm when you identitfy $G$ with $U(G)$.
By the way, Definition 2' is conceptually not complete yet. You have to assume $x \cdot i(x) = i(x) \cdot x = e$. This becomes important when you study group objects in non-cartesian categories, aka Hopf monoids, for example Hopf algebras. The axiom for the antipode then contains two diagrams.
Best Answer
No. Let $X$ be any finite set with at least two elements, and consider the monoid $\mathrm{End}(X)$, of functions $X \to X$, under composition.
Each constant function $c$ is a left-absorbing element, i.e. $cf = c$ for any $f$. However, there are no right-absorbing elements in the monoid. So $\mathrm{End}(X)$ cannot be isomorphic to its opposite, since any such isomorphism would have to interchange left- and right-absorbing elements.