The answer is no, if the Fredholm index of $L$, which is the integer
$\mathrm{ind}\, L=\dim\mathrm{ker}\, L- \dim \mathrm{ker} \, L^\ast$,
is negative. (Regard $L$ as an unbounded operator on $L^2(M)$ with domain $D(L)=H^{2k}(M)$.) A proof by contradiction goes as follows. If the eigenvectors formed a basis of $H^k(M)$, then their span would also be dense in $L^2(M)$. The elements of $\mathrm{ker}\, L^\ast$, i.e. of the null space of the adjoint operator, are orthogonal to the range of $L$, which contains the eigenvectors with non-zero eigenvalues. The assumption $\mathrm{ind}\, L<0$ implies that there exists a non-zero element of $\mathrm{ker}\, L^\ast$ orthogonal to *all* eigenvectors. Thus we obtain the contradiction that the span of the eigenvectors is not dense in $L^2(M)$.

I have no explicit example of a scalar elliptic operator of even order with negative index. Because of $\mathrm{ind}\, L^\ast= -\mathrm{ind}\, L$ it would suffice to find an operator with non-zero index. By the Atiyah-Singer index theorem, this is a topological condition on the manifold $M$ and on the principal symbol of the operator $L$.

You received the answer "yes" to your question on SE under the additional assumption that $L$ is self-adjoint and coercive. It was explained there how this follows using the spectral theorem and using, what I am accustomed to call, Gelfand triplets. As a reference for the latter I mention the book: Wloka, *Partial Differential Equations*. Note that self-adjointness implies that the index is zero.