Are all finite-dimensional algebras of a fixed dimension over a field isomorphic to one another?

Another very familiar example: $\mathbb{C}\neq\mathbb{R}\times \mathbb{R}$. The complex numbers are a field, but $(1,0)(0,1)=(0,0)$ in $\mathbb{R}\times \mathbb{R}$, so it has non-trivial zero-divisors.

They will not necessarily be isomorphic. Consider $V = \mathbb F[x] / (x^n)$ and $W = \mathbb F^n$ with componentwise multiplication.These are both $n$ dimensional $\mathbb F$ algebras. However, $V$ contains a nilpotent element, $x$, whereas $W$ contains no nilpotent elements. Indeed, if we had an $\mathbb F$-algebra homomorphism $f: V \longrightarrow W$ then as $0 = f(x^n) = f(x)^n$, we'd need $f(x) = 0$ so any map between the two must have a nontrivial kernel.

In general, the answer is "no", even if one requires $V$ and $W$ to be fields.

For example, the rings $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are two non-isomorphic fields that both have dimension $2$ over $\mathbb{Q}$.