If a PDE has a unique classical solution, must it have a unique viscosity solution?

The answer in general is no, because viscosity solutions are not suitable for every problem.

The viscosity solution idea based on an idea of solution comparison. If you look at the user's guide, Crandall-Ishii-Lions 1992, you see that for a problem of the form $$ F(x,u,Du,D^2u)=0 $$ what they call the fundamental assumption is that $$ F(x,r,p,X)\leq F(x,s,p,Y) \mbox{ whenever } r\leq s \mbox{ and } Y\leq X. $$ There must be a way to define subsolutions and supersolutions. You cannot do this for an arbitrary PDE. For example, if you consider an oscillatory problem, e.g. Maxwell's Equations, that seems very difficult.

On the other hand, if your problem is of the correct type, then yes, of course. A basic references to know what correct type means are the User's guide given above. I would also look at Guy Barles' book (1994), where he gives a number of conditions grouped under the idea of 'Natural conditions' (e.g., coercivity, convexity etc..) where uniqueness holds for parabolic problems.

For a parabolic equation, yes.

The reason is that viscosity solutions are essentially defined to have comparison with classical solutions.

Here is the proof. Call the viscosity solution $u$ and the classical solution $\phi$. Suppose that they are not equal, so assume for example that $u-\phi$ is positive somewhere, say $(x,t) \in U \times (0,\infty)$. Then (assuming that they have the same initial and Dirichlet boundary condition on $\partial U$) the function $u-\phi$ achieves a positive maximum in $U \times (0,t]$. Then for $\epsilon > 0$ small enough, if we set $\tilde\phi(y,s):= \phi(y,s) + \epsilon s$, then $u - \tilde \phi$ achieves a positive maximum at some point of $U \times [0,t]$. This is a contradiction to the fact that $u$ is a viscosity subsolution (since $\tilde \phi$ is a strict, smooth supersolution).