Conjugacy classes in the fundamental group

Use Lemma 1.19 in that very book. It says that if $\varphi_t:S^1\to X$ is a homotopy and if $h$ denotes the path $\varphi_t(s_0)$ formed by the images of $s_0$ then $\varphi_{1*}$ is equal to the composition $$\pi(S^1,s_0)\xrightarrow{φ_{0*}}\pi(X,φ_0(s_0))\xrightarrow{\beta_h}\pi(X,φ_1(s_0))$$ where $\beta_h([f])=[\overline h\cdot f\cdot h]$. In the case $φ_0(s_0)=φ_1(s_0)=x_0$ the path $h$ becomes a loop. Now given loops $f$ and $g$ at $x_0$, the equality $\Phi([f])=\Phi([g])$ implies a free homotopy $φ_t$ from $φ_0=f$ to $φ_1=g$ such that $φ_t(s_0)$ is a loop $h$. The formula for $\beta_h$ then gives the conjugacy. The other direction should not pose difficulties considering that you knew how to show the surjectivity.

Sometimes it is easier to prove a more general result. Let $Y$ be a space with well pointed base point $y$, i.e. $(Y, \{ y \})$ has the HEP, and let $X$ be a space with base point $x$. Consider the map of homotopy classes $p: [Y,X] _\bullet \to [Y,X]$ where the former is the base point preserving homotopy classes and the latter is the free homotopy classes. The result is that if $X$ is path connected then $p$ is surjective and the group $\pi_1(X,x)$ operates on the set $[Y,X]_\bullet$ so that the quotient is $[Y,X]$.

This is actually 7.2.12 of Topology and Groupoids; the proof there uses the notion of fibration of groupoids, which is fun anyway, but the key hint is that you need the HEP to get the operation given above. I hope that helps.