The circle with centre $(a,0)$ and radius $r$ is tangent to $x=0$ if $r=|a|$. It is then tangent to $y=e^x$ and $y=-e^x$ if there is $x$ such that $e^{2x} + (a-x)^2 = a^2$ and $2 e^{2x} + 2(x-a) = 0$. Eliminating $a$, the equation for $x$ is $(1-2x) e^{2x} - x^2 = 0$. This has two real solutions, approximately $-.7478435353725761$ and $.4580318389459811$, corresponding to $a = -.5237489523800181$ and $2.957464268427510$ respectively. I don't think there are "closed-form" solutions, even with the use of the Lambert W function.
Once you have the solution for the first circle, similar methods will get further circles. Thus the circle with centre at $(a_2, 0)$ and radius $r_2$ is tangent to $y=e^x$ and $y=-e^x$ and $x = 2 a$ (and thus to the first circle) if $r_2 = |a_2 - 2 a|$ and $e^{2x} + (a_2 - x)^2 = r_2^2$ and $2 e^{2x} + 2 (x - a_2) = 0$ For the case $a = -.5237489523800181$, I get $a_2 = -1.308841381001648$.
Here's a picture.
