It follows that $e_k(n) = \begin{bmatrix} n+1\\ n-k+1\end{bmatrix}$, i.e., unsigned Stirling numbers of first kind.

Now, let
$$f_n(x) := \sum_{k\geq 0} e_k(n) x^k = (1+x)(1+2x)\cdots (1+nx).$$
Then
$$\sum_{k\geq 0} e_{2k}(n) x^{2k} = \frac{1}{2}(f_n(x)+f_n(-x))$$
and
$$a_+(n) = \frac{1}{2}(f_n(I)+f_n(-I)) = \Re f_n(I),$$
where $I$ is the imaginary unit.

Let $r:=n\bmod 3$ and $m:=\lfloor n/3\rfloor$, and so $n=3m+r$. Then
$$f_n(x) \equiv ((1+x)(1+2x)(1+0x))^m \prod_{j=1}^r (1+jx) \equiv (1+2x^2)^m \prod_{j=1}^r (1+jx)\pmod{3}.$$
Correspondingly,
$$a_+(n) \equiv (-1)^m \prod_{j=1}^r (1+jI) \equiv
\begin{cases} (-1)^m &\text{if}\ r=0,1\\
(-1)^{m+1} &\text{if}\ r=2\end{cases} \pmod{3}.$$

This can be combined into
$$a_+(n) \equiv (-1)^{\lfloor (n+1)/3\rfloor}\pmod{3}$$
and so $a_+(n)\equiv \lambda_{n+1}\pmod3$ (modulo the typo in the definition of $\lambda_n$).