Can a function share an interval with another, different function? I'm not quite sure how to put this into words, but I'm really curious. Could you have two (defined, continuous non-piecewise) functions intersect "over an interval"? 
To put that another way, if you picked a continuous interval, say 0 to 1, and knew the value of a function f(x) for any x along that interval, could you correctly determine a UNIQUE general function that would define f(x) for ALL x, that was the ONLY (again, continuous and non-piecewise) function that satisfied all x for your initial interval?
 A: Non-piecewise is not really a meaningful property of a function. A particular description of a function may or may not be by cases, but the function itself is just the correspondence between input and output values; it does not remember how you choose to define it.
Differentiable, as suggested in a comment, is better -- at least that is an actual property of a function. In that case, two differentiable functions can certainly share values on on interval and be different elsewhere, such as
$$ \begin{align} & f(x) = 0 \\ & g(x) = \frac{(\sqrt{x^2}-x)x}2 = \begin{cases} x^2 & \text{when } x \le 0 \\ 0 &\text{otherwise} \end{cases} \end{align}$$
However, I suspect that the concept you really want (but don't know) is analytic. An analytic function is one that is differentiable arbitrarily often and equals its Taylor series in a neighborhood of every point. And it is true that two analytic functions on $\mathbb R$ that agree on an interval must agree everywhere.
(Most total functions that you can write down without resorting to case splits will automatically be analytic, so it is not completely crazy to suggest "non-piecewise" as a criterion. It is just not easy at all to make a rigorous concept out of it. For example, the first expression for $g(x)$ above contains the trick $|x|=\sqrt{x^2}$, and there are examples of functions defined with integrals that aren't even continuous, despite not obviously being "by cases").
It is not enough to require that the function is smooth, that is, differentiable arbitrarily often. The standard counterexample to that is
$$ \begin{align} & f(x) = 0 \\ & g(x) = \begin{cases} 0 & \text{when } x \le 0 \\ e^{-1/x} &\text{when } x>0\end{cases} \end{align}$$
(It may not be obvious that $g$ is differentiable arbitrarily often at $0$, but it is true).
Polynomials are always analytic so it is true that two polynomials that agree on an interval agree everywhere. In fact, two polynomials that agree at infinitely many points (no matter how distributed) must agree everywhere.
