The graph of $S$ as a function of $x$ looks very much like a hyperbola, which suggests using the approximation $S(x,\lambda)\approx a+\frac{b}{x+c}$ where $a,b,c$ depend on $\lambda$ in some way.
I first tried to fit a hyperbola to the data $S(0)=\frac{\lambda}{1-\lambda},\quad S'(0)=-\frac{\lambda(1+2\lambda)}{(1-\lambda)^2}, \quad S(1)=\frac{1}{2\lambda^2}\left(2\lambda-(1-\lambda^2)\ln\frac{1+\lambda}{1-\lambda}\right)$ by solving the system $a+b/c=S(0)$, $-b/c^2=S'(0)$, and $a+b/(1+c)=S(1)$. Unfortunately this results in a substantial underestimate. The reason is that $S'(0)$ is a poor indicator of the behavior of the curve: $S'$ changes rapidly near $0$.
Then I fit the model $a+b/(x+c)$ to three points $S(0), S(1/3), S(1)$, and this worked much better.
Here is the comparison for $\lambda=0.17$ ($S$ in red, approximation in blue):

For $\lambda=0.37$

For $\lambda=0.87$

The coefficients $a,b,c$ are ugly, so I did not reproduce them here. The ugliness does not matter much if you work with specific $\lambda$. If you also need reasonably simple dependency on $\lambda$, let us know: maybe someone will come up with a better way to get $a,b,c$.
(Added)
Here is an approximation with simple dependency on $\lambda$. I found it more convenient to work with $\tilde S=1-\frac{1-\lambda}{\lambda}S$, since this function is increasing from $0$ to something in the range 0.3..1, which is to say it has a milder dependency on $\lambda$ than the original one. I picked the model $\tilde S\approx \alpha x/(\beta x+\gamma)$ where $\alpha,\beta,\gamma$ are linear functions of $\lambda$. Here is one possibility: $ \tilde S\approx \frac{(1+2\lambda)x}{(2+\lambda)x+1-\lambda} $ This approximation is not great, but you can tweak the six (well, actually five) numeric parameters to make it look better.