A heuristic argument based on Ofer's comment
Note that $T := YY^\top = XX^\top + AX^\top +X^\top A + AA^\top$, where $A:=\alpha 1_n 1_d^\top$ is the all-ones matrix of shape $n \times d$. Note that $AA^\top=\alpha^2 d1_n1_n^\top$.
Fact 1: $\mbox{LSD}(XX^\top + AX^\top +X^\top A + AA^\top) \to \mbox{LSD}(XX^\top + AA^\top)$, in some sense (probably weakly).
The above fact (it seems) was implied by Ofer's comment. I've observed this empirically, but I'm not quite sure why it should be true. Maybe some due to some free probability lemma ?
Also, from Lemma 1 of this monograph, we deduce that
Fact 2 (Golub's formula): There exist $q_1,\ldots,q_n \ge 0$ with $\sum_{i=1}^n q_i = 1$, such that $\lambda_i(XX^\top+AA^\top) = \lambda_i(XX^\top + \alpha^2d 1_n1_n^\top) = \lambda_i(XX^\top) + q_i\alpha^2nd$.
Moreover, we have the formula $q_i = \dfrac{(1_n^\top u_i)(1_n^\top v_i)}{nu_i^\top v_i}$ (provided $u_i^\top v_i \ne 0$), where $u_1,\ldots,u_n$ are the eigenvectors of $XX^\top$ and $v_1,\ldots,v_n$ are the eigenvector of $v_i$. Intuitively, one whould expect $u_i$ to be pretty much orthogonal to $v_i$ for $i \ne 1$. As a consequnce, one would expect $q \to (1,0,\ldots,0)$, in some sense.
Putting things together, one would expect $\lambda_i(YY^\top) \to \lambda_i(XX^\top) + \delta_{i=1}\alpha^2 nd$.