The inequality $\frac{\alpha + \lambda_{max}(A)}{\alpha+\lambda_{min}(A)} > \frac{\lambda_{max}(A)}{\lambda_{min}(A)}$ doesn't hold because $\lambda_{max}(A) > 1 > \lambda_{min}(A) > 0$ and $\alpha > 0$. In fact, the reverse inequality holds in this case. Hence, there is no contradiction here.

The inequality $\frac{\alpha + \lambda_{max}(A)}{\alpha+\lambda_{min}(A)} > \frac{\lambda_{max}(A)}{\lambda_{min}(A)}$ doesn't hold because $\lambda_{max}(A) > 1 > \lambda_{min}(A) > 0$ and $\alpha > 0$. In fact, the reverse inequality holds in this case. Hence, there is no contradiction here.

EDIT: From the comments, I realized that the question is not answered completely yet.

Also, $\kappa(A+\alpha I) = \frac{\alpha + \lambda_{max}(A)}{\alpha+\lambda_{min}(A)} = \frac{\frac{\alpha}{\lambda_{max}(A)}+1}{\frac{\alpha}{\lambda_{max}(A)}+\frac{\lambda_{min}(A)}{\lambda_{max}(A)}} = \frac{\frac{\alpha}{\lambda_{max}(A)}+1}{\frac{\alpha}{\lambda_{max}(A)}+O(h^2)}$.

Hence, for small enough $h$ and big enough $\alpha$, we have $\kappa(A) \approx 1+\frac{\lambda_{max}(A)}{\alpha}$. One can see that $\lambda_{max}(A)$ changes very little. For example,

poisson = @(n) full(gallery('poisson',n,n));
max(abs(eig(poisson(10))))

ans =

7.837971894457977

max(abs(eig(poisson(40))))

ans =

7.988263204734964

max(abs(eig(poisson(60))))

ans =

7.994696359539318