Expression for fugacity coefficient derived from a pressure-explicite EOS
Mindful of the guideline "don't give answers in comments", I've converted my comment to an answer, trivial though it is.

When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.

When you write down the integrated form, $z-\ln z$, you need to remember that it means that you are evaluating the function $z(v)$ at the two limits $v=\infty$ and $v=v$, and using those values in that expression; you should not be trying to set $z=v$ at the limits, which seems to be (partly) what you have written at the end of your question.

Edit following OP comment.

More details. At constant $T$, $z=z(v)$ is a function of molar volume $v$,
determined by the equation of state, $P(v)$ at the given value of $T$. This equation is unknown, but we can be sure that,
in the ideal gas limit $v\rightarrow\infty$, $z(v)\rightarrow 1$.

Integrating both sides of the equation from the ideal gas limit to 
the desired volume $v$, and using $v'$ as the integration variable, gives
eqn (2) of the question in its full form:
$$
\left| \ln \phi(v')\right|_{v'=\infty}^{v'=v} 
= \left|  z(v') - \ln z(v') \right|_{v'=\infty}^{v'=v} 
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v' 
$$
I think it's clearer to distinguish between the integration variable $v'$
and the "upper" limit of integration $v$, 
but many people would be happy just to use $v$ instead of $v'$.

Anyway,
on the left,
we know that, in the ideal gas limit, 
the fugacity coefficient $\phi(\infty)=1$,
so $\ln\phi(\infty)=0$ and we are just left with $\ln\phi(v)$.
On the right,
similarly,
we substitute in the upper and lower limits for $v'$.
We know $z(\infty)=1$, so 
the function being evaluated is $z(\infty)-\ln z(\infty)=1$ at the lower limit,
and $z(v)-\ln z(v)$ at the upper limit.
So the final answer is
$$
\ln \phi(v)
= z(v) - \ln z(v) - 1
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v' 
$$
An important point is that the integration variable 
is the molar volume $v'$ (or $v$ if you prefer), not $z$.
In evaluating the result, the integration limits $v$ and $\infty$ 
are substituted for $v'$, the argument of the function being evaluated.
It is incorrect to set $z=v$, or $z=\infty$.
(This should be even more clear in this case, 
since $z$ is a dimensionless quantity, whereas $v$ is not).