It is my first instinct to assume this is true, but I'm not sure my work is sufficient to prove it. We begin with a convergent series $\sum^\infty_{n=1} a_n$ with positive, real terms. We want to determine if $\sum^\infty_{n=1}(e^{a_n}-1)$ is also convergent. My first idea was to first expand the exponent. This gives us
$$ \sum^{\infty}_{n=1} \sum^{\infty}_{k=0}\left(\frac{a_n^k}{k!}-1\right).$$ We can then "peel" off the zeroth term of the $k$ series giving us $\frac{a_n^0}{0!} = 1$. This cancels with the -1. Now, we are left with,
$$\sum^{\infty}_{n=1} \sum^{\infty}_{k=1}\left(\frac{a_n^k}{k!}\right) .$$ Since we know $a_n$ converges, let's call its limit $a$. Now,
$$ \sum^{\infty}_{k=1}\left(\frac{a^k}{k!}\right) < \sum^{\infty}_{k=0}\left(\frac{a^k}{k!}\right) = e^a .$$ Thus we can conclude $\sum^\infty_{n=1}e^{a_n}-1$ is less than $e^a$ which is finite, thus it too must be finite.
Alternatively, one can use the limit comparison test. That is, for two series $\sum s_n$ and $\sum a_n$ with positive terms,
$$ \lim_{k\rightarrow\infty} \frac{s_k}{a_k} = \rho.$$
If $\rho >0$ then both series converge or both diverge. In this case we have,
$$ \lim_{n\rightarrow\infty}\frac{a_n} {e^{a_n}-1} = \frac{a}{e^a-1}$$ since our terms are positive and nonzero, clearly $\rho > 0$ thus both must converge.
Best Answer
If $(a_n)$ are assumed to be positive, your conjecture is true. If there is no other assumption, it's wrong.
Counterexample $a_n= \frac{(-1)^n}{\sqrt{n}}$.
Indeed, $\sum a_n$ is convergent, however $$\sum e^{a_n}-1 \stackrel{(*)}{\ge}\sum a_n+\frac{a_n^2}{2e} = +\infty$$ where in $(*)$ I have used the fact that $e^x \ge 1+ x+ \frac{x^2}{2e}$ for all $x \ge -1$