Is the set of quaternions $\mathbb{H}$ algebraically closed?
If by $K[x]$ you mean the algebra freely generated by $K$ and an indeterminate $x$, then — as noted in my comment — $p(x)=ix+xi-j$ has no root in $\Bbb{H}$, because $ix+xi$ always lies in the plane spanned by $\{1, i\}$.
On the other hand, if by $K[x]$ you mean the subset of that free algebra consisting of expressions of the form $\sum k_i x^i$, we can generalize a topological proof of the fundamental theorem of algebra, as follows.
Theorem: Let $K$ be a finite-dimensional normed $\Bbb{R}$-algebra with $Z(K)=\Bbb{R}$, such that the subalgebra generated by each non-central element is isomorphic to $\Bbb{C}$. Then for any $k_0,\dots,k_{n-1} \in K$, there exists an $x \in K$ such that $x^n+\sum_{i=0}^{n-1} k_ix^i=0$.
Proof: Let $g(x)=x^n$; let $S(K)$ and $B(K)$ be the unit sphere and unit ball in $K$. Since the subalgebra generated by each non-central element is isomorphic to $\Bbb{C}$, every element of $S(K)$ except $\pm 1$ has $n$ preimages under $g$. A lengthy but straightforward Jacobian computation shows that $g$ is orientation-preserving at its regular values; thus the restriction of $g$ to $S(K)$ has topological degree $n$.
Now, suppose for the sake of contradiction that $f(x)=x^n+\sum_{i=0}^{n-1} k_ix^i$ is never zero, and let $f_t(x)=t^nf(x/t)=x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i$. Then $f_t$ is also nonvanishing. Define a map $\gamma_t:B(K) \to S(K)$ by $\gamma_t(x)=\dfrac{f_t(x)}{|f_t(x)|}$. Since $B(K)$ is contractible, the restriction of $\gamma_t$ to $S(K)$ has topological degree $0$.
But $\gamma_t=\dfrac{x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i}{\left|x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i\right|}$ is homotopic to $g$; since topological degree is a homotopy invariant, we have a contradiction.
This proves exactly the statement you were looking for when $K=\Bbb{H}$, and for monic polynomial functions of this form in general.
When $K=\Bbb{O}$, this doesn't quite prove the statement you were looking for. Given an arbitrary polynomial function in the form you want, you can't necessarily divide through by the leading coefficient to get a monic polynomial function of the form you want, because of non-associativity. It's not too hard to adapt, though; for any $\omega \in \Bbb{O}$, left-multiplication by $\omega$ is a nonsingular linear map, so $g(x)=\omega x^n$ has nonzero degree and the rest of the proof still goes through.
$K=\Bbb{S}$ has zero-divisors, so the statement is trivially false there.
Eilenberg and Steenrod, in Foundations of Algebraic Topology, prove a form of algebraic closure for both the quarternions and the octononions (ch. 11, sec. 5). The proof uses topological degree theory. The polynomial is required to be of the form $m+g$, where $m$ is a monomial of degree $n$ and $g$ is a sum of monomials all of degree less than $n$. However, monomials like $axbxc$ (of degree 2) are allowed.
One can be a bit more general: if by $K[x]$ is meant the algebra of generalised polynomials,
Note. A division ring $K$ of finite dimension $[K:k]>1$ over its centre $k$ is never algebraically closed.
Proof. For every $a\in K\setminus k$, being $k$-linear of Kernel having dimension $\geqslant 1$, the polynomial $x\mapsto ax-xa$ is non-surjective by the Rank-nullity Theorem, which gives rise to degree $1$ rootless polynomials.