Infinite-dimensional normed division algebras

A MathSciNet search reveals a paper by Urbanik and Wright (Absolute-valued algebras. Proc. Amer. Math. Soc. 11 (1960), 861–866) where it is proved that an arbitrary real normed algebra (with unit) is in fact a finite-dimensional division algebra, hence is one of the four mentioned in the OP. A key piece of the argument (Theorem 1) is to show that such an algebra $A$ is algebraic, in the sense that if $x \in A$, then the subalgebra of $A$ generated by $x$ is finite-dimensional. The authors then invoke a theorem of A. A. Albert stating that a unital algebraic algebra is a finite-dimensional division algebra.

Hurwitz's theorem is stated here (section 2.6 of 'A taste of Jordan algebras' by Kevin McCrimmon) as:

Any composition algebra $C$ over a field $\Phi$ of characteristic not 2 has finite dimension $2^n$ for $n=0,1,2,3$ and is one of the following ...

it goes on then to describe generalisations of the usual normed division algebras.

The wikipedia page composition algebra tells us that one only has a 1-dimensional composition algebra when the characteristic of the base field is not 2, but otherwise you can start from a 2-dimensional composition algebra over a characteristic 2 field and perform the usual Cayley-Dickson construction.

Edit: The following theorem was proved by Kaplansky (Infinite-dimensional quadratic forms admitting composition, Proc AMS 1953), which finishes off the classification. A quadratic form $g$ on an algebra $A$ over a field $F$ in this context is a function $g:A \to F$ such that $g(kx) = k^2g(x)$ for $k\in F$ and $x\in A$.

Theorem. Let $A$ be an algebra with unit element over a field $F$. Suppose that $A$ carries a nonsingular quadratic form $g$ satisfying $g(xy) = g(x)g(y)$ for all $x, y \in A$. Then:
(a) A is alternative,
(b) except for the case where $A$ has characteristic two and is a purely inseparable field over $F$, $A$ is finite dimensional and of dimension 1, 2, 4, or 8,
(c) $A$ is either simple or the direct sum of two copies of $F$,
(d) $g(x) = x^\ast x$ where $x\mapsto x\ast$ is an involution of $A$.

So unless your base field has characteristic 2, and your division algebra is a purely inseparable extension of the base field, your division algebra has to be finite dimensional.

The associative case follows from Mazur's Theorem (see here). He proved that there are up to isomorphism precisely three Banach division algebras, namely $\mathbb R,\mathbb C$ and $\mathbb H$. This applies to the completion of any normed division algebra, which still verifies the identity $|ab|=|a||b|$, and hence is a division algebra.