Good question indeed. 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 $$ R\,E(z,s+1) = (s+1)\,yE_2(z,s).$$ 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 (principal series) 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 $$ (s+1)-\frac{1}{2}=s+\frac{1}{2}\qquad\text{and}\qquad \frac{1}{2}-(s+1)=-s-\frac{1}{2}.$$
Well, we still need to specify all this to $s=0$: the archimedean Langlands parameters in this case equal $1/2$ and $-1/2$.
Added. Actually, $E(z,s+1)$ belongs to the unitary principal series only when $\operatorname{Re}(s)=-1/2$. The OP is really interested in the case $s=0$, in which case the corresponding automorphic representation belongs to the limit of discrete series (denoted by $\mathcal{D}^\pm(1)$ in Bump's book "Automorphic forms and representations", see page 216 there). This added section is based on David Loeffler's remarks below, for which I am grateful.