Last question first: degree is not (depending on how you look at it) a property of elements of polynomial rings. It's a property of elements of graded polynomial rings, and the easiest way to choose a grading is to choose a set of generators. Your example is a little confused: when you consider $k[x^2]$ you are implicitly considering $x^2$ to have degree $1$, but you don't need to adopt this convention; you can declare that it has degree $2$ instead. (In any case, the notation $k[x^2]$ is misleading: when you write this you are really talking about the entire inclusion map $k[x^2] \to k[x]$, so everything depends on whether you want this to be a map of rings or a map of graded rings.)
Your point 3 seems to answer your first question. I'm not sure why you find $k[\mathbb{A}^n]$ objectionable but $k[V]$ not given that I assume you think $\mathbb{A}^n$ is a variety. I think this is a fine way to refer to a polynomial ring without naming its generators. Your worry about the distinction between polynomials and the functions they induce can be ignored if $k$ is algebraically closed, and otherwise you should just remind yourself that in the remaining cases the functor from $k$-varieties to $\text{Set}$ isn't faithful, so the answer is not to take the set-theoretic picture too seriously in the first place and work directly with the opposite of the category of finitely-generated reduced $k$-algebras. In this category I describe exactly how to recover the ring of functions geometrically in this blog post. (I should be more explicit about what I mean here: if you work in the right category, the polynomial ring in $n$ variables over $k$ is the space of functions on $\mathbb{A}^n$.)
I also don't understand your first two sentences; they seem to be inconsistent with each other. (Off-topic: I don't know what you look like, but because of my Gravatar you know what I look like. I'm the guy sitting in the back of class on his laptop, so if you'd like to introduce yourself that would be cool.)
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
Define a commutative ring structure on $\mathbb Z$ as follows: $a+'b=p(a,b)$ and $a\times'b=q(a,b)$, where $p(X,Y),q(X,Y)\in\mathbb Z[X,Y]$. Note that since $p$ and $q$ are symmetric, we have that $p,q\in \mathbb Z[X+Y,XY]$.
For each $a\in\mathbb Z$ we must have $\mathbb Z\to\mathbb Z:x\mapsto x+'a=p(x,a)$ is bijective, hence $p(X,a)$ must be a degree-$1$ polynomial of $X$. Thus $p(X,Y)$ must be of the form $\alpha+\beta(X+Y)+\gamma XY$ for $\alpha,\beta\in\mathbb Z$. Now we need the map $\mathbb Z\to\mathbb Z:x\mapsto x+'a=\alpha+\beta a+(\beta+\gamma a)x$ to be bijectice for each $a\in\mathbb Z$, so that $\gamma=0$ and $\beta=\pm1$. However, if $\beta=-1$ we must have that
$$p(p(X,Y),Z)=p(\alpha-X-Y,Z)=X+Y-Z$$
is symmetric with respect to permutation of $X,Y,Z$, which is clearly not true. Thus we conclude that $\beta=1$ and $p(X,Y)=\alpha+X+Y$. Thus, $0'=-\alpha$, and by shifting everything by $\alpha$ (that is, applying the bijection $\mathbb Z\to\mathbb Z:x\mapsto x+\alpha$) we may simply assume $\alpha=0$.
Now, $q$ must be a polynomial such that $q(X+Y,Z)=q(X,Z)+q(Y,Z)$, hence it must be of the form $q(X,Y)=f(X)Y$ for some polynomial $f(X)\in\mathbb Z[X]$. Since $q$ must be symmetric, it must be the case that $f(X)=aX$ for $a\in\mathbb Z\setminus\{0\}$. The existence of a multiplicative unit tells us that $a=\pm1$. Either way, by considering the isomorphism $\mathbb Z\to\mathbb Z:x\to ax$ we may assume $a=1$, so that $(\mathbb Z,+',\times')\cong(\mathbb Z,+,\times)$.
P.S.
We may arrive at the same conclusion without assuming commutativity, since distributivity on the left tells us that $q(X,Y)$ is of the form $Xg(Y)$ for some $g(Y)\in\mathbb Z[Y]$.