Prove that the following statements are equivalent for a nonzero ring D:
(i) $D$ is a division ring.
(ii) For all $ a, b ∈ D $ with $ a \neq 0 $, the equations $ ax = b $ and $ ya = b $ have unique solutions in $ D $.
(iii) For all $ a, b ∈ D $ with $ a \neq 0 $, the equation $ ax = b $ has a solution in $ D $.
(iv) $ D^{2} \neq 0 $ and $D$ has no right ideals other than $0$ and $D$.
Suppose $D$ is unital and associative. We can show TFAE:
$\rm(i)$. $D$ has two-sided multiplicative inverses for nonzero elements.
$\rm(ii)$. The equations $ax=b$ and $xa=b$ are always solvable when $a\ne0$.
$\rm (iii)$. The equation $ax=b$ is always solvable when $a\ne 0$.
$\rm (iv)$. $D$ has no zero divisors and no nontrivial proper right ideals.
$\rm (i)\Rightarrow(ii)$: Given $ax=b$ we may left multiply by $a^{-1}$ to get $x=a^{-1}b$, and similarly we may right multiply $xa=b$ by $a^{-1}$ to get $x=ba^{-1}$, so the equations are solvable.
$\rm (ii)\Rightarrow(iii)$: This is obvious.
$\rm (iii)\Rightarrow(iv)$: Suppose $J$ is a nontrivial right ideal of $D$. Pick $a\in J$ nonzero. Then $ax=1$ is solvable, but we know $ax\in J$, so $1\in J\Rightarrow J=D$.
$\rm (iv)\Rightarrow(i)$. Suppose $a\in D$ is nonzero and consider the right ideal $aD$. Since $aD\ne 0$, we know we must instead have $aD=D$, and thus $ax=1$ for some $x$. Left multiply by $x$ to get $(xa)x=x$, which gives $(xa-1)x=0$, which since there are no zero divisors gives $xa=1$ and so $x$ is in fact a two-sided inverse.
Without the assumption of associativity, things go haywire. To begin with, I'm not sure if ideals really make sense, so we'll throw that condition out.
The algebra above showing a right inverse is a left inverse doesn't use associativity, but it does use flexibility, a special case of alternativity. But without flexibility, it is possible for an element to have distinct left and right inverses.
Example A. Consider the "perturbed" quaternions $(\mathbb{H},\bullet)$ in which the basis elements $1,\mathbf{i},\mathbf{j},\mathbf{k}$ have the same multiplication table except $\mathbf{i}^2=-1$ is replaced by $\mathbf{i}\bullet\mathbf{i}=-1+\varepsilon \mathbf{j}$.
One may rewrite $\mathbf{j}$ as either $\mathbf{k}\bullet\mathbf{i}$ or $-\mathbf{i}\bullet\mathbf{k}$, collect terms, factor $\mathbf{i}$ out of the appropriate side, and move the minus sign to obtain either of
$$ \begin{cases} (-\mathbf{i}+\varepsilon\mathbf{k})\bullet\mathbf{i}=1 \\ \mathbf{i}\bullet(-\mathbf{i}-\varepsilon\mathbf{k})=1 \end{cases} $$
Thus, $\mathbf{i}$ has distinct left and right multiplicative inverses with respect to $\bullet$.
On the other hand, the $\bullet$ operation has no zero divisors. Write the operation as
$$\mathbf{x}\bullet\mathbf{y}=\mathbf{xy}+\varepsilon x_1y_1\mathbf{j}.$$
(where $\mathbf{xy}$ is usual quaternion multiplication). Set this equal to $0$, move $\varepsilon x_1y_1\mathbf{j}$ to the other side, then take norms of both sides assuming $x_1y_1\ne 0$ and $|\varepsilon|<1$ to get a contradiction.
For finite rings $R$ or finite-dimensional algebras over fields, the absence of zero divisors is equivalent to $ax=b$ and $xa=b$ always being solvable. Both are equivalent to the bijectivity of the left-multiplication-by-$a$ and the right-multiplication-by-$a$ maps (exercise). However, I don't know if these are equivalent otherwise.
Example B. Not only may a ring have no zero divisors but have distinct left/right multiplicative inverses, it may have zero divisors even though two-sided multiplicative inverses always exist. This is the case with the 16-dimensional number system called the sedenions.