Find $a$ from the minimum distance between $y=e^x$ and $y=\\log_e x=\ln x$

The two curves are inverses of each other, so they are symmetric about the line $y=x$.

Their intersection points satisfy $e^x=\ln x.$ It is easy to check that $x=1$ gives $e^1=e \neq 0,$ so that is not an intersection. Instead, note that for inverse curves, the minimum distance occurs on the line $y=x$ when the corresponding points coincide. Thus we solve $e^x=x.$ But this has no real solution.

So let us use the standard result for inverse curves: the minimum distance between $y=f(x)$ and $y=f^{-1}(x)$ is attained at their intersection point(s), and if they do not intersect, we minimize directly.

Consider points $(t,e^t)$ on $y=e^x$ and $(e^t,t)$ on $y=\ln x$ by symmetry. Their distance is $D=\sqrt{(e^t-t)^2+(t-e^t)^2}=\sqrt{2}\,|e^t-t|.$ So minimize $|e^t-t|.$ Since $\phi(t)=e^t-t,$ we have $\phi'(t)=e^t-1=0 \implies t=0.$ Then $\phi(0)=1.$ Hence minimum distance is $a=\sqrt{2}.$ Therefore, $a^2=2.$

Eccentricity of the ellipse

Given $x^2+a^2y^2=25a^2.$ Divide by $25a^2$: $\frac{x^2}{25a^2}+\frac{y^2}{25}=1.$

Since $a^2=2>1$, we have $25a^2=50>25,$ so the major axis is along $x$.

Thus $A^2=25a^2,\qquad B^2=25.$ The eccentricity of the ellipse is $e_1=\sqrt{1-\frac{B^2}{A^2}}=\sqrt{1-\frac{25}{25a^2}}=\sqrt{1-\frac1{a^2}}.$ With $a^2=2$, $e_1=\sqrt{1-\frac12}=\frac{1}{\sqrt2}.$

Eccentricity of the hyperbola

Given $x^2-a^2y^2=5.$ Write as $\frac{x^2}{5}-\frac{y^2}{5/a^2}=1.$ So $A^2=5,\qquad B^2=\frac{5}{a^2}.$

The eccentricity of the hyperbola is

Use the condition: ellipse eccentricity is $b$ times hyperbola eccentricity

Given $e_1=b e_2.$ Hence $b=\frac{e_1}{e_2}=\frac{1/\sqrt2}{\sqrt{3/2}}=\frac1{\sqrt3}.$ Therefore, $b^2=\frac13 \implies \frac1{b^2}=3.$

Compute the required expression

$a^2+\frac1{b^2}=2+3=5.$

Conclusion

The required value is $\boxed{5}.$ So the correct option is D.