Tauberian theorem $\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} $

We may take for a counterexample any convergent series $\sum_{k=1}^\infty c_k$ with $\sum_{k=1}^\infty |c_k|^2<\infty$ and $\delta_k:={1\over kc_k}\to0$. Take $\lambda_k:=k(1-\delta_k)$, so that ${\lambda_k\over k}\to1$. Then for any $t>0$ $$\sum_{k\ge1} e^{-\lambda_k t}c_k-\sum_{k\ge1} e^{-k t}c_k=\sum_{k\ge1} e^{-k t}\,{e^{\delta_k kt}-1\over {\delta_k kt }}\,t\ge\sum_{k\ge1} e^{-kt-|\delta_k|k t}\;t=$$$$= \int_0^{+\infty}e^{-x}dx+o(1)=1+o(1)\, ,$$ because ${e^u-1\over u}\ge e^{-|u|}$ for all $u\neq0$, and because $\delta_k=o(1)$ ). Taking into account the result you listed at point (4) we have $$\liminf_{t\to0_+} \sum_{k\ge1} e^{-\lambda_k t}c_k\ge1+\sum_{k=1}^{\infty}c_{k}\, . $$

Let me try one more time, inspired by Pietro's idea.

Let's for now focus on $t=1/N$ and an interval $I_N$ of length $N$ located near $k\simeq (N/3)\log N$. Let's take $\lambda_k=(1\pm\delta)k$ on this interval, with $\delta=1/\log N$, and the sign is the same as that of $c_k$. Then the terms of $$ \sum_{k\in I_N} c_k(e^{-\lambda_k t}- e^{-kt}) = \sum e^{-k/N}c_k(e^{\pm\delta k/N}-1) \quad\quad\quad\quad (1) $$ are non-negative. In fact, let's also take $c_k$ with alternating signs. Then the sum is $\gtrsim N^{-1/3} \sum |c_k|$. We can easily make this large, for example by giving $|c|$ the constant value $|c_k|=N^{-1/2-\epsilon}$ on $I_N$ (note that this will keep $\sum_{I_N} |c_k|^2$ small, as required by the $\ell^2$ condition).

Now we define the whole sequence by first choosing $N_j$'s that increase very rapidly, and then defining $c_k$ as above on each of these intervals, and $c_k=0 $ otherwise. Notice that for $t=1/N_j$, if $k$ is taken from one of the other intervals $I_m$, $m\not= j$, then $e^{-\lambda_kt}-e^{-kt}$ is extremely small, because $tk$ is either very small or very large. So these intervals make negligible contributions to (1).

It follows that (1) does not go to zero as $t\to 0+$ along the sequence $t=1/N_j$, and as discussed earlier, this means that we have a counterexample.

Finally, an obvious modification also gives examples where $c\in\ell^p$ for any (or all) $p>1$.

The $\lambda_k$ correspond to the Laplacian eigenvalues of a domain in $\mathbb{R}^2$ implies that \begin{equation} 0<\lambda_1\le \lambda_2\le\cdot\cdot\cdot\le \lambda_n\le\cdot\cdot\cdot \end{equation} and $\lambda_n\rightarrow\infty $ as $n\rightarrow\infty$. Hence for any $t>0$ \begin{align} \sum_{n=1}^{\infty}e^{-\lambda_nt}c_n&=\sum_{n=1}^{\infty}[C(n)-C(n-1)]e^{-\lambda_nt}\\ &=\sum_{n=1}^{\infty}C(n)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\\ &=C(\infty)+\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right], \end{align} by Abel transform, where $$C(n)=\sum_{k=1}^{n}c_k ~\mbox{for}~ n\ge 1 $$ and $$o_n(1)=\sum_{k=1}^{\infty}c_k-C(n)=C(\infty)-C(n)\rightarrow 0~as~ n\rightarrow \infty.$$ Then it is obviously that $$\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\rightarrow 0~ as~t\rightarrow 0+.$$

In fact, for any $\varepsilon>0$, there exist an $N_{\varepsilon}\in\mathbb{N}$ such that for all $n>N_{\varepsilon}$, $|o_n(1)|< \varepsilon$ holds. Furthermore, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \left|\sum_{n=1}^{N_{\varepsilon}}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|+\epsilon\sum_{n>N_{\varepsilon}}\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right].$$ Namely, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \max_{1\le k\le N_{\varepsilon}}|o_k(1)|e^{\lambda_1t}+\epsilon.$$ The next let $t\rightarrow 0$ and then let $\varepsilon\rightarrow 0$, then the results is obvious.