Seems that it can not have more than $n$ roots. Let us use the following generalization of Descartes rule for signed measures. Namely, let $\mu$ be a Borel signed measure on a real line, with compact support (this condition may be of course weakened). We say that $\mu$ has at most $k$ changes of sign if there exist points $c_1<c_2<\dots <c_k$ such that restriction of $\mu$ to each of the intervals $(-\infty,c_1],(c_1,c_2],(c_2,c_3],\dots,(c_k,\infty)$ is either non-negative or non-positive. I claim that in this situation Laplace transform $L(t):=\int e^{tx} d\mu(x)$ has at most $k$ real roots with multiplicity counted (except the case $\mu\equiv 0$). Induction in $k$. Base $k=0$ is clear. Assume that $k\geq 1$, for $k-1$ it is established, we should prove for a measure $\mu$ having exactly $k$ sign changes. If $L(t)$ has $N$ roots, $g(t):=e^{c_1t}(L(t)e^{-c_1t})'$ has $N-1$ roots, but $g(t)=\int (x-c_1)e^{tx} d\mu(x)$, and new signed measure $d\nu(x)=(x-c_1)d\mu(x)$ has at most $k-1$ sign changes (if new measure $\nu$ is 0, then $\mu$ was proportional to the delta-measure in $c_1$, and $L(t)$ does not have roots at all.) So, by induction hypothesis $N-1\leq k-1$, as we need.
Now we say $e^{at}-e^{bt}={\rm sign}\,(a-b)\cdot t\cdot \int_{[a,b]} e^{tx} dx$, so if we divide our equation $\sum_{i=1}^n e^{a_it}-e^{b_it}=0$ by $t$, it becomes a Laplace transform for a measure with density $\sum \pm \chi_{[a_i,b_i]}$, which has at most $n-1$ sign changes. Thus our equation has at most $n-1$ non-zero roots as desired.
(PREVIOUS VERSION, working for simple roots) It is equivalent to counting positive roots of $\sum x^{a_i}=\sum x^{b_i}$. There is always root 1, so subtract $n$ from both parts and divide by $x-1$. I hope that after this the difference has at most $n-1$ sign changes and so Descartes rule is applicable (if we assume that exponents are positive integers, as we may assume: having more than $n$ roots is open condition, so exponents may be considered rational, then by scaling change of variables integer.) Indeed, sign change appears if there were more $a$'s than $b$'s on some initial segment, but it became more $b$'s than $a$'s. It requires two points at least.
UPD: when I say that having more than $n$ roots is an open condition, I, strictly speaking, lie. It is so for simple roots, but for multiple roots it is not: say, small perturbations of $x^2$ may have no real roots. However I believe that the statement is true for multiple roots too, but it have to be fixed somehow.
