This is not true. Let's take $\lambda_k=k$. The problem is that for a given small $t>0$, there is a long intermediate interval where $e^{-kt}$ is no longer close to $1$, but not yet small enough to make $\sum |e^{-kt}c_k|$ negligible, and we can build a sequence $c_k$ to order that exploits this.

Let's focus on $t=1/N$ and $N\le k\le N\log N$. On the first half of this interval, let's take $c_k=\delta N^{-1/2}$, with a small $\delta>0$ to be determined later, and $c_k=-\delta N^{-1/2}$ on the second half of this interval. Then
$$
\sum_{k=N}^{N\log N} e^{-k/N}c_k = e^{-1}\delta N^{1/2} + O(\delta) ,
$$
which of course is a disaster since $\sum_{k=N}^{N\log N} c_k = 0$.

We now just take a very rapidly increasing sequence $N_j$ and $\delta_j=1/j$, say, and then define $c_k$ as above on the corresponding intervals on $c_k=0$ otherwise. Then for $t=1/N_j$, we have $\sum e^{-kt} c_k \simeq N_j^{1/2}/j\not\to 0=\sum c_k$: for $N_j\le k\le N_j\log N_j$, the discussion above applies, and for the other intervals, $e^{-k/N_j}$ is either very close to $1$ or very small, so these sums are extremely close to zero.

Finally, note that $c\in\ell^2$, as desired; in fact, an obvious modification gives us examples with $c\in\ell^p$ for any (or all) $p>1$.