How can I compute $$ \prod_p \left(1+\frac{k}{p}\right)\exp(-k/p) $$ where $0<k<e$ and the product is over all primes $p$?
Background
L. G. Sathe proved [1] that there are $$ (1+o(1))f\left(\frac{\nu}{\log\log x}\right)\frac{x}{\log x}\frac{(\log\log x)^{\nu-1}}{(\nu-1)!} $$ squarefree integers up to $x$ with $\nu$ prime factors, where $\nu$ grows as a fixed multiple of $\log\log x$ (constant from 0 to $e$). This is similar to Landau's earlier theorem which omits $f$ and fixes $\nu$. The definition of $f$ is $$ f(k)=\frac{C^k}{\Gamma(k+1)}\prod_p\left(1+\frac kp\right)\exp(-k/p) $$ with $$ C=\prod_p\frac{p-1}{p}\exp(1/p)=0.7292647442571190188536\ldots $$
Note that MR0056625 misquotes the formula for $f$ above.
[1] L. G. Sathe, On a problem of Hardy on the distribution of integers having a given number of prime factors. I., J. Indian Math. Soc. (N.S.) 17 (1953), pp. 63-82.
Taking the logarithm, we see that
$$\begin{align} \log \prod_{p > k} \left(1 + \tfrac{k}{p}\right) \exp \left(- \tfrac{k}{p}\right) &= \sum_{p > k} \log \left(1+\tfrac{k}{p}\right) - \tfrac{k}{p}\\ &= \sum_{p > k} \sum_{n=2}^\infty (-1)^{n-1} \frac{k^n}{n\cdot p^n}\\ &= \sum_{n=2}^\infty (-1)^{n-1} \frac{k^n}{n} \underbrace{\sum_{p > k} \frac{1}{p^n}}_{g_{k}(n)}. \end{align}$$
The change of order of summation is legitimate by the absolute convergence.
Since $\frac{k^n}{n}g_k(n)$ is monotonically decreasing in $n$, we have an alternating series, and thus any two consecutive partial sums give bounds for the total sum, and the error is always smaller than the first omitted term.
For $k < 2$, we have $g_k(n) = \zeta_p(n)$, the prime zeta function, but we can also uniformly work with $g_2(n) = g_e(n) = \zeta_p(n) - 2^{-n}$ and multiply with $\left(1+\frac{k}{2}\right)\exp (-k/2)$ after exponentiating the computed approximation to the logarithm. It may however be more efficient to uniformly work with $g_m$ for larger $m$ and treat the primes $\leqslant m$ separately, since then the convergence of $\frac{k^n}{n}g_m(n)$ to $0$ is faster.
So altogether we have
$$\prod_p \left(1+\tfrac{k}{p}\right)\exp \left(-\tfrac{k}{p}\right) = \left[\prod_{p\leqslant m}\left(1+\tfrac{k}{p}\right)\exp \left(-\tfrac{k}{p}\right)\right] \exp \left(\sum_{n=2}^\infty (-1)^{n-1}\frac{k^n}{n}g_m(n) \right)$$
for arbitrary $m \geqslant k$. Choosing a larger $m$ leads to fewer $g_m(n)$ needed to achieve the desired accuracy, but computing
$$g_m(n) = \zeta_p(n) - \sum_{p \leqslant m} \frac{1}{p^n}$$
to the desired accuracy becomes more work the larger $m$ is. As a guesstimate, I'd think some $m \in [5,19]$ would give the best trade-off.