How does a Lewis acid differ from an oxidizing agent; Lewis bases from reducing agents?
An acid and an oxidising agent are very different concepts that apply to different phenomena — although a single compound can be both at the same time, e.g. $\ce{H2SO4}$. However, a compound can also be an acid and a reducing agent, e.g. $\ce{HI}$.

The Brønsted acid/base theory is clearly distinct from any type of redox reaction, so I suppose that the confusion lies with the extension of said theory that the Lewis theory provides. But first a short word on the Brønsted theory: therein, an acid is a compound that can transfer a $\ce{H+}$ ion onto another compound.

The Lewis theory realises, that the acidic transfer reagent need not be hydrogen. Instead, anything with an unoccupied low-energy orbital can function as a Lewis acid; most notably boron compounds and metal cations. Upon creation of a Lewis acid-base pair, a coordinate bond is formed. Basically, this means that the energy of the lone pair of the Lewis base is lowered because it interacts with the empty orbital of the Lewis acid whose energy is raised. This can be depicted as $\ce{LB\bond{-&gt;}LA}$. The key concept is that the interaction is fully reversible and that upon dissociation the Lewis base keeps its lone pair.

When discussing oxidising or reducing agents, however, the electrons are irreversibly transferred. They actually move from orbitals of the reducing agent (which will be oxidised) into orbitals of the oxidising agent (which will be reduced). This transfer is thermodynamically favourable and thus the electrons won’t just ‘go back’.

To exemplify this, let’s consider two reactions of $\ce{H2SO4}$; one where it is an oxidising agent and one where it is an acid (Brønsted type; the proton of $\ce{H2SO4}$ is actually the Lewis acid).

$$\text{Acid-base:}\\
\ce{H2SO4 (l) + NH3 (g) &lt;=&gt;&gt; NH4+ HSO4- (s)}$$

Note that ammonia, our base, is a gaseous compound at room temperature. If you employ appropriate conditions (i.e. temperature and low pressure), the equilibrium will be drawn to the left, releasing ammonia as ammonia ($\ce{NH3}$ molecules) including its lone pair. $\ce{H2SO4}$ will stay behind.

$$\text{Oxidising agent:}\\
\ce{2H2SO4 + Cu -&gt; CuSO4 + SO2 ^ + 2 H2O}$$

$\ce{SO2}$, being a gas, will diffuse away. However, even if you perform this in a pressurised vessel, there will be no significant back-reaction because the electrons have been fully transferred from copper to sulfur.



The argument for Lewis bases/reducing agents is the same in reverse.