Can $\sum_{x \in [0,1]} e^x$ be represented as an integral?

Since nobody has wrote about this yet, let me mention that this sum indeed represents an integral, but not the integral you thought of. Namely, for any set $X$ and any function $f:X\to\Bbb R$ $$ \sum_{x\in X}f(x) = \int_Xf(x)\mu(\mathrm dx) $$ where $\mu$ is the counting measure. In your case $X = [0,1]$, and you instead thought of integrating with a more natural Lebesgue measure $\lambda$ which is of course different from the counting one, so $$ \sum_{x\in[0,1]} f(x) = \int_{[0,1]}f(x)\mu(\mathrm dx)\neq \int_{[0,1]}f(x)\lambda(\mathrm dx) $$ in general. In particular, the inequality does not hold for $f(x) = \exp (x)$ as other answers showed.

In fact, both integrals only agree (being finite) if they both are zero. Indeed, if the counting integral is non-zero and finite, then $f$ takes only countably many non-zero values - but then the Lebesgue integral is zero. Vice-versa, if the Lebesgue integral is non-zero, then $f$ takes uncountably many values so that either the counting integral is undefined, or infinite.

No, they're not equal; the sum of uncountably many positive numbers is necessarily infinite, whereas the integral $$\int_0^1\exp(x)\,dx=e-1$$ is finite.

Not exactly. Taken literally, $\sum_{x\in[0,1]}\exp(x)$ is a sum with uncountably many positive summands and cannot evaluate to a finite value. (For example, thee are more than $10^{100}$ summnds and all are $\ge1$, hence the sum is at least $>10^{100}$).