After unsuccessfully trying some more substitutions and hoping that they would look like $1-\frac1{something}$ I decided to take the direct approach. Calling the original quantity $f(x)$ I defined $f(x)=1-\frac1{g(x)}$ and calculated a Taylor series for $g(x)=\frac1{1-f(x)} = \frac{\log x}{\log x - \log (x-1)}$ . The derivatives were messy but fortunately evaluating at $x=2$ removes a lot of terms due to the $\log (x-1)$ terms.
To first order $g(x) = 1 + (x-2)\frac1{\log 2}$ which when inverted leads to $x=2-\log 2 + \frac{\log 2}{1-f(x)}$ . This meets my criteria of estimating x large (approximating $f(x)$ small); it is a fair bit off but does provide a bound. The second order approximation is much closer but unfortunately the series appears to take alternating signs past there and thus second order estimates to the wrong side of the function. Of course third order would be pretty accurate but very messy to invert.
Are there any techniques for adjusting the second order function to be guaranteed smaller than the original function? (Hmm - I just remembered that I like Rational Polynomials because you can get an extra degree of freedom while keeping linear functions - might have to work through that approach.)
I might have to expand around $x=5$ or so to get a closer match at first order and just accept numerical coefficients instead of exact expressions. Or I can give in and do a lookup table or search with the forward function. Since I'm only interested in the values of the original function at the primes that is feasible - I always knew that would work but wanted a one-step calculation.
Update
In the course of writing the preceding I recalled the handy properties of Pade Approximants so I generated some first order expressions. I found that I got better results by choosing three points to send the function through than expanding around a single point using derivatives. Given that success I decided to go for it and generated a ratio of quadratics with numerical solutions. Sending the function through the values at 2, 3, 5, 7, and $\infty$ I got $f(x) = \frac{3.025072 + .861544 \cdot x - 1.18704 \cdot x^2}{1 + .547267 \cdot x - 1.18704 \cdot x^2}$
This function is very close to the true value. I have not attempted to verify that it is always conservative to the function, but a numeric check shows this to be the case through $300$ which is enough for my purposes. The sharp reader may note that this fit exceeds the function at the reals $\lt 7$ (and at $4$ and $6$) but since I am only interested in the values at the primes this is not a problem.