This result of Euler is the last theorem in http://eulerarchive.maa.org/docs/originals/E072.pdf, and if you prefer English to Latin look at the last theorem in http://eulerarchive.maa.org/docs/translations/E072en.pdf. I don't know why Sandifer says the proof can't be repaired. Euler's goal is to show $\sum_{p} 1/p$ diverges, and that in some sense $\sum_{p \leq n} 1/p$ diverges like $\log(\log n)$. (Euler didn't speak directly about asymptotic approximations, and he wrote the last conclusion as $\sum_p 1/p = l(l(\infty))$, using $l$ for the natural logarithm.)
Euler's proof: for each positive integer $k$, set $A_k = \sum_p 1/p^k$. Then
$$
e^{\sum_{k \geq 1} A_k/k} = \sum_{n \geq 1} \frac{1}{n}
$$
and $\sum_{k \geq 2} A_k/k < \infty$, so $A_1$, which is $\sum_p 1/p$, diverges.
Since $A_1 = \infty$ while $\sum_k A_k/k < \infty$, we have
$$
e^{A_1} = \sum_{n \geq 1} \frac{1}{n},
$$
so
$$
\sum_p \frac{1}{p} = A_1 = \log\left(\sum_{n \geq 1} \frac{1}{n}\right).
$$
QED
The last part is kind of a bogus calculation, but the rest can be fixed using the zeta-function.
In place of $A_k$ use $\sum_{p} 1/p^{ks}$, so the replacement of the first displayed equation above is
$$
e^{\sum_{p,k} 1/kp^{ks}} = \sum_{n \geq 1} \frac{1}{n^s}
$$
for ${\rm Re}(s) > 1$. In other words, this is saying $\sum_{p,k} 1/kp^{ks}$ is a logarithm of $\zeta(s)$ for ${\rm Re}(s) > 1$. Series are absolutely convergent, so all manipulations of terms in the series are justified. Euler does all his work at $s = 1$, where expressions may be divergent.
Euler's claim that $\sum_{k \geq 2} A_k/k < \infty$ is correct, and the modern replacement for this is the more general observation that $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$. Euler's statement is at $s=1$.
Let $s$ be real and greater than 1, so the displayed equation can be rewritten as
$$
\sum_{k \geq 1} \sum_{p} \frac{1}{kp^{ks}} = \log\left(\sum_{n \geq 1} \frac{1}{n^s}\right).
$$
As $s \rightarrow 1^+$, $\sum_n 1/n^s \rightarrow \infty$, so the right side above tends to $\infty$ as $s \rightarrow 1^+$. On the left side above, split off the terms where $k=1$ from the terms where $k \geq 2$. Since $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$, this series has a limit as $s \rightarrow 1^+$ (Dirichlet series are continuous on open half-planes where they converge). Therefore the sum of the terms with $k = 1$, $\sum_p 1/p^s$, must tend to $\infty$ as $s \rightarrow 1^+$. For $s > 1$ we have $\sum_p 1/p > \sum_p 1/p^s$, and the right side tends to $\infty$ as $s \rightarrow 1^+$, so $\sum_p 1/p = \infty$.
