Excellent question indeed. The quick answer is that $E_2(z)$ is an almost holomorphic modular form of weight $2$, so it does not belong to an automorphic representation in the usual sense of the word. For more details (and my thought process), read below.
Consider the Maass raising operator $$ R:=y\left(i\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right).$$ Let $(m,n)\in\mathbb{Z}^2$ be a nonzero pair of integers. Then a small calculation gives that, for $z=x+iy$, $$ R\left(\frac{y^s}{|mz+n|^{2s}}\right) =\frac{sy^s}{(mz+n)^2|mz+n|^{2s-2}}.$$
Now let us introduce the usual weight $0$ level $1$ (nonholomorphic) Eisenstein series $$ E(z,s):=\sum_{\substack{m, n \in \mathbb{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{|mz + n|^{2s}},$$ then we see that $$ R\,E(z,s+1) = (s+1)\,yE_2(z,s).\tag{$\ast$}$$ On the right hand side, $yE_2(z,s)$ is the canonical weight $2$ level $1$ (nonholomorphic) Eisenstein series, the one which transforms as a weight $2$ Maass form. It is worthwhile to recall here that weight $k$ holomorphic forms embed into the weight $k$ Maass spectrum by multiplying each weight $k$ holomorphic form by $y^{k/2}$. In our case $k=2$, which explains why we multiply by $y$.
So your Eisenstein series, after inserting the factor $y$ to make it into a canonical weight $2$ form, and also insertig the scaling factor $s+1$, equals the Maass raising shift of $E(z,s+1)$. It belongs to the same automorphic representation as $E(z,s+1)$, hence it has the same Langlands parameters as $E(z,s+1)$ at every place. In particular, the archimedean Langlands parameters are $$ (s+1)-\frac{1}{2}=s+\frac{1}{2}\qquad\text{and}\qquad \frac{1}{2}-(s+1)=-s-\frac{1}{2}.$$
Added and revised. Well, we still need to specify all this to $s=0$, but in this case the above argument breaks down, because $E(z,s+1)$ does not exist in this case (it has a pole at $s=0$). So we need to be more careful. Let us use the results and notation of Section 4 of Duke-Friedlander-Iwaniec: The subconvexity problem for Artin L-functions (Invent. Math. 149 (2002), 489-577). Then for $\Re s>1$ we have the Fourier decomposition \begin{align*} \frac{1}{2}E(z,s)&=\ \zeta(2s)y^s+\pi^{2s-1}\frac{\Gamma(1-s)}{\Gamma(s)}\zeta(2-2s)y^{1-s}\\&+\ \frac{\pi^s}{\Gamma(s)}\sum_{n=1}^\infty\frac{\sigma_{2s-1}(n)}{n^s}\bigl\{f_0^+(nz,s)+f_0^-(nz,s)\bigr\}.\end{align*} Let us replace $s$ by $s+1$ here, and then apply the raising operator $R$ along with $(\ast)$. Then for $\Re s>0$ we obtain the Fourier decomposition \begin{align*} \frac{1}{2}E_2(z,s)&=\ \zeta(2s+2)y^s+\pi^{2s+1}\frac{\Gamma(1-s)}{\Gamma(2+s)}\zeta(-2s)y^{-s-1}\\&-\ \frac{\pi^{s+1}}{y\Gamma(2+s)}\sum_{n=1}^\infty\frac{\sigma_{2s+1}(n)}{n^{s+1}}\bigl\{f_2^+(nz,s+1)+s(s+1)f_2^-(nz,s+1)\bigr\}.\end{align*} The right hand side is indeed holomorphic at $s=0$, and at this value it specifies to \begin{align*} \frac{1}{2}E_2(z)&=\ \frac{\pi^2}{6}-\frac{\pi}{2y}- \frac{\pi}{y}\sum_{n=1}^\infty\frac{\sigma_1(n)}{n}f_2^+(nz,1)\\ &=\ \frac{\pi^2}{6}-\frac{\pi}{2y}-4\pi^2\sum_{n=1}^\infty\sigma_1(n)e(nz).\end{align*} I hope I got everything right. At any rate, it is clear now that the $L$-function of $E_2(z)$ is $\zeta(s-1)\zeta(s)$, and $E_2(z)$ should belong to the holomorphic discrete series of weight $2$ even though its constant term is not holomorphic. I think this paradox arises from the fact that $E_2(z)$ is not a true automorphic form.
Added 2. Indeed, $E_2(z)$ is an almost holomorphic modular form of weight $2$. See Section 2.3 of Bruinier-v.d.Geer-Harder-Zagier's book "The 1-2-3 of modular forms", in which $G_2^*(z)$ is precisely our $\frac{1}{2}E_2(z)$ above. In particular, (19) and (21) reveal that $E_2(z)$ indeed transforms precisely like a weight $2$ holomorphic modular form (i.e. without any correction terms), even though it is not holomorphic. One can learn more about almost holomorphic modular forms in Section 5.3 of the book.