Here's the way I would do it, using only algebra and basic properties of the exponential function, with no (explicit) use of calculus: start by setting $d_i = a_i \exp(-b_i c_i)$ for $i = 1, 2, 3$. Then your equation becomes
$d_1 \exp(b_1 x) + d_2 \exp(b_2 x) = d_3 \exp(b_3 x)$.
In this equation we can safely assume that $b_1 \ge b_2 \ge b_3$; since the $a_i$ may have any sign, we can re-arrange/re-name them any way we like. Now assume not all the $b_i$ are equal. Then we may write
$d_1 + d_2 \exp((b_2 - b_1)x) = d_3 \exp((b_3 - b_1)x)$.
Now if $b_1 > b_2, b_3$, so that $b_2 - b_1, b_3 - b_1 < 0$, we let $x \to \infty$ and conclude that $d_1 = 0$, hence $a_1 = 0$ and we obtain
$d_2 \exp((b_2 - b_1)x) = d_3 \exp((b_3 - b_1)x)$,
and multiplying through by $\exp(b_1x)$,
$d_2 \exp(b_2x) = d_3 \exp(b_3x)$.
Next we observe that either $d_2 = d_3 = 0$, or $d_2 \ne 0 \ne d_3$, since neither exponential factor can ever vanish. The former case corresponds to the trivial "solution"
$a_1 = a_2 = a_3 = 0$; in the latter case we have
$d_2 = d_3\exp((b_3 - b_2)x)$,
which in turn forces $b_2 = b_3$ since otherwise the left-hand side is constant and the right-hand side is not! Of course then $d_2 = d_3$ as well.
Having dealt with the case $b_1 > b_2 \ge b_3$, and remembering our hypothesis that $b_1 \ge b_2 \ge b_3$, it only remains to deal with the case $b_1 = b_2 > b_3$. Then we find that
$d_1 \exp(b_1 x) + d_2 \exp(b_1 x) = d_3 \exp(b_3 x)$,
whence
$d_1 + d_2 = d_3 \exp((b_3 - b_1)x)$,
and again the fact that $d_1 + d_2$ is constant but $d_3 \exp((b_3 - b_1)x)$ isn't creates a contradiction with the assertion $b_1 = b_2 > b_3$ unless of course $d_1 + d_2 = d_3 = 0$.
A brief summary of what has been accomplished so far, remembering that $b_1 \ge b_2 \ge b_3$, and excluding the case $a_1 = a_2 = a_3 = d_1 = d_2 = d_3 = 0$:
Case I: $b_1 > b_2 \ge b_3$: $d_1 = a_1 = 0$, $b_2 = b_3$, $d_2 = d_3$;
Case II: $b_1 = b_2 > b_3$: $d_1 + d_2 = d_3 = 0$;
Case III: $b_1 = b_2 = b_3$: $d_1 + d_2 = d_3$.
Under these circumstances the case $b_1 > b_2 > b_3$ is ruled out under Case I, since it forces $b_2 = b_3$. And of course in Case III we still have $a_1\exp(b_1c_1) + a_2\exp(b_1c_2) = a_3\exp(b_1c_3)$ which allows the values of the $a_i$ and $c_i$ to be traded off against each other to a certain extent. Similar trade-offs apply in some of the other cases as well; I leave it to the reader to discover these.
The key point here is that, unless the coefficients are very specifically constrained,
the $b_i$ must all be the same.
I believe these results generalize to sums of more than two exponentials. The key to the analysis, without calculus, I believe, is to divide out by the factor $\exp(b_ix)$ with the largest (or smallest) $b_i$, and use properties of the $\exp(b_ix)$ as $x \to \pm \infty$ to force the values of certain of the other constants, as we did with the $a_i$, $d_i$.
The reason I said, at the beginning of this answer, is that there is "no (explicit) use of calculus" is that there is no differentiation, integration, etc., deployed in this answer.
Of course, it could reasonably be argued that the definition of $\exp(b(x - c))$ itself involves arguments based on continuity, limits, and other topological properties of the real number system; but these are not quite calculus in my book.
To make more general progress, and/or to analyze similar statements for other functions, say $\sin(bx)$ etc., may very well require the machinery of differentiation, Wronksians and Vandermonde determinants, as Andre Nicolas mentioned in his comment.
Best Answer
Use the Inverse function theorem. The jacobian of $f$ is $e^x\ne0$. This gives local invertibility. To show that $f$ is not a injective on $\mathbb{R}^2$, consider $f(x,y)$ and $f(x,y+2\,k\,\pi)$, $k\in\mathbb{Z}$.