To answer my own question, I think the quote I gave from the book is confused to the point of being nonsensical.
I will use the discussion from the Wikipedia article on the 2nd Law of Thermodynamics, under the heading of "Energy, available useful work". I am adapting and simplifying the argument. See Wikipedia for a fully rigorous treatment, including issues I will leave out (chemical energy, difference between work and useful work).
We imagine the system in contact with an infinite reservoir, whose pressure and temperature can be assumed to remain constant. Let $S$ be the entropy of the system, $S_R$ be the entropy of the reservoir, and $S_{TOT}$ be the total entropy. By the 2nd Law, we have $$dS_{TOT}=dS+dS_R \ge 0 \Rightarrow dS \ge -dS_R \tag{1}$$ By the 1st Law, we have for the system $$dU=dQ-dW \tag{2}$$ where $U$ is its internal energy, $dQ$ is the heat flowing into the system and $dW$ is the work done by the system. Now $$dQ=T(-dS_R) \le T dS$$ where $T$ is the temperature of the reservoir, and we use Eqn 1. Substituting for $dQ$ in Eqn 2 and rearranging, we get $$dW \le -dU + TdS \tag{3}$$ Now we define the 'exergy' of the system (I think this is energy available for doing useful work) by $$E=U - T\,S \Rightarrow dE=dU-TdS$$ Eqn 3 can therefore be rewritten as $$dW \le -dE$$ $$\Rightarrow dW+dE \le 0 \tag{4}$$ Now if we consider work done on the system, $dW'$, we have $dW=-dW'$, so Eqn 4 becomes $$-dW'+dE \le 0$$ $$\Rightarrow dE \le dW'$$ or in other words, "the system cannot receive more in energy [available for work] than the amount of external work supplied". QED.
This is the path by which we go from "$dS \ge 0$" (meaning $dS_{TOT} \ge 0$) to this statement, which was my original question.
Having gone through this, I think the confusions of the book are as follows:
1. Uses $S$ to represent both total entropy and system entropy, and does not distinguish between them.
2. 749
Writes the equality $dS = \mu\,dE + \mu\,dW$, which should really be an inequality $dS \ge \mu\,dE + \mu\,dW$ (a rearrangement of Eqn 3 above, with $\mu=1/T$, and the symbol $E$ in place of $U$).
3. Uses $E$ to represent internal energy, instead of the usual $U$, and fails to make clear that, in the statement about "energy" received being less than work applied, this really refers to exergy or available energy, and is not the same as the $E$ in the equation.
Edited to add...
In fact, I now see a simpler way of looking at this. The basic problem is that $\mu$ in the original equation is negative, and this is concealed.
Imagine there is no heat transfer, so it is just a question of work done. If, as usual, $dW$ represents work done by the system. We have, by the 1st Law, $dU=-dW$. With the definition of exergy, $E$, above, $dE=dU-T\,dS$, this becomes $dE+T\,dS=-dW$ or, rearranging, $T\,dS=-dE-dW$, which is the same as the author's equation, with $\mu=-1/T$. Now if we have work done on the system, $dW=-dW'$, we can write $T\,dS=-dE+dW'$, and if $dS \ge 0$, this becomes $dW'-dE \ge 0$, which means $dW' \ge dE$, i.e. the useful energy received is less than the work done on the system, QED. I think this must be basically what the author had in mind. 749