It looks like the answer is morally "no." Now, there is a formal closure for which we can solve free polynomials: As in the case of fields, you take an inductive limit. Let $\Bbb H=R$ be our normed division algebra. Then

$$\overline{R}=\varinjlim_{[L:R]<\infty} L$$

where the inductive system is taken relative to inclusions of algebra extensions $L/R$ of finite dimension over $R$, each of the form

$$L_p=R\{x\}/(p(x))$$

where $p(x)$ is irreducible over $R$ and $R\{x\}$ is the polynomials freely generated as in case ($2$). This certainly has the required property that all polynomials in $R$ have a root in $\overline{R}$, and any other such object has a copy of this inside of it for purely formal reasons.

I note that the directed system so-defined is indeed a directed system--in fact a lattice--so this should go through unless I'm missing something obvious.

rschweib has noted that the result is no longer a division algebra, so this is really not ideal, but the "algebraic closure" property holds, and necessarily it's a minimal ring where this property can hold, so it seems this is the best we can hope for. However we **also** cannot force algebraicness of the result since $R\{x\}/(xi+ix-j)$ doesn't make $x$ algebraic appropriately in the sense that you want to mimic the field case's excellent definition that algebraicness means $F(\alpha)/F$ is finite dimensional as an algebra over $F$, which doesn't hold in this setting.