Your argument purports to show the chemical potential is positive. You look it up and find that it's zero. In fact, the chemical potential for an ideal gas is negative!
First, I'm not following the step where you conclude that the entropy must go down because the pressure goes up. The entropy of an ideal gas is given by the Sackur-Tetrode equation:
$$
S=Nk\left[\ln\left({V\over N}\left(4\pi m U\over 3Nh^2\right)^{3/2}\right)+{5\over 2}\right]
$$
Incidentally, I got this particular form of the equation from the textbook by Schroeder, which is the one I like best for this sort of thing.
Under the circumstances you describe (increasing $N$ by 1 while holding $U/N$ constant), this reduces to
$$
S=Nk\left[\ln(V/Nv_Q)+ {5\over 2}\right],
$$
where $v_Q$, the quantum volume per particle, is constant and $Nv_Q\ll V$. (The latter is the condition for the gas to be nondegenerate.) The derivative of this with respect to $N$ is positive: upon adding a particle at constant $T,V$, the entropy goes up by
$$
\Delta S=k\left[\ln(V/nv_Q)+{3/2}\right]>0.
$$
Second, where are you looking up the value for $\mu$ for argon at STP? Is it possible that the table you're looking in lists values relative to STP? It's certainly not true that the chemical potential of an ideal gas at STP is zero.
One way to get the chemical potential of an ideal gas is by imagining adding the new particle in such a way that the total energy (not the temperature) remains fixed. The relevant identity then is
$$
\mu=-T\left(\partial S\over\partial N\right)_{U,V},
$$
which comes from $dU=T\,dS-P\,dV+\mu\,dN$. Taking the derivative of the above expression for $S$, we get
$$
\mu=-kT\ln\left[{V\over N}\left(2\pi mkT\over h^2\right)^{3/2}\right].
$$
Again, for gases that are far from degenerate, the quantity inside the logarithm is large, so the chemical potential is negative.
According to the first law of thermodynamics
\begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align}
Where $Y$ is the generalized force, $dX$ is the generalized displacement.
Helmholtz Free Energy
\begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align}
Gibbs Free Energy
\begin{align}G=U-TS-YX=\sum_j\mu_jN_j.\end{align}
Therefore that
\begin{align}G=F-YX.\end{align}
In your case, $Y=H$, $X=M$, so we get
\begin{align}G=F-HM.\end{align}
You can see the textbook:
A Modern Course in Statistical Physics by L. E. Reichl, 2nd, ed (1997), p23, 42, 45.
Best Answer
A thermodynamic potential $P$ satisfies $$P(\lambda X, Y) = \lambda P(X,Y)$$ where $X$ are extensive and $Y$ are intensive parameters of state. As an example $P$ could be the internal energy of a system, it then doubles if we double all its extensive properties (volume, particle number, etc).
If we take $\lambda = \tfrac{1}{N}$, we get $$P(\tfrac{X}{N}, Y) = \tfrac{1}{N} P(X,Y)$$ Let us denote the intensive versions (such as energy per particle, volume per particle) of all extensive quantities with lower case letters: $x=\tfrac{X}{N}$, the same also applies for the potential: $p(X,Y)=\tfrac{1}{N}P(X,Y)$. We then get: $$P(x,Y) = \tfrac{1}{N} P(X,Y) = p(X,Y)$$.
Let us write this down for the Gibbs free energy $G=G(T,p,N)$. Note that its only extensive dependency is $N$, the other two, $p$ and $T$ are intensive. $$G(T,p,\tfrac{N}{N}) = G(T,p,1) = \frac{1}{N}G(T,p,N) = g(T,p,N)$$
Obviously, since the l.h.s does not depend on $N$, also the r.h.s. doesn't, so: $g=g(T,p)$. So we can rewrite the last step of the above equation as: $$G(T,p,N) = Ng(T,p)$$ Diffferentiating with respect to $N$ gives:
$$\left(\frac{\partial G}{\partial N}\right)_{T,p} = g(p,T)$$
On the other hand, from the total differential of $G$ that is given as $dG=-SdT+Vdp+\mu dN$ we know that $$\left(\frac{\partial G}{\partial N}\right)_{T,p} = \mu$$
Therefore, in total, we get: $$G(T,p,N) = Ng(T,p) = N\mu$$
Note that from this derivation you can follow Euler's relation. Indeed, since $G$ is defined as Legendre transform of $U$, $$G=U-TS+pV$$ we have, by plugging in $G=\mu N$ from above, $$N\mu = U-TS+pV$$ or $$U=TS-pV+N\mu$$