Let's start by defining some notation.
$$
y_i\in\{-1,+1\}\\
z_i\in\{0,1\}
$$
Then $z_i = (y_i+1)/2 \iff 2z_i = y_i + 1 \iff y_i = 2z_i-1$.
Also, $p(y_i\equiv1) = p(z_i\equiv1)$ and $(y_i\equiv-1) = p(z_i\equiv0)$.
Also:
$$p(y_i\equiv1) = p(z_i\equiv1)= \left(1+\exp(-w^Tx_i)\right)^{-1}\\
p(y_i\equiv-1) = p(z_i\equiv0)= \left(1 + \exp(w^Tx_i)\right)^{-1}$$
When $y_i = -1$, then $\dfrac{y_i + 1}{2} = 0$ and $\dfrac{y_i - 1}{2} = -1$. When $y_1 = +1$, then $\dfrac{y_i + 1}{2} = 1$ and $\dfrac{y_i - 1}{2} = 0$. Consequently:
$$
\color{red}{\log\left(1 + \exp(-y_i w^Tx_i)\right)}\\
=\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i-1}{2}\right)\log\left(1 + \exp(w^Tx_i)\right)
$$
Then $1-2 = -1$, so we get:
$$
\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i-1}{2}\right)\log\left(1 + \exp(w^Tx_i)\right)\\=
\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i+(1-2)}{2}\right)\log\left(1 + \exp(w^Tx_i)\right)
$$
For the fraction on the right, $\dfrac{y_i+(1-2)}{2} = \dfrac{y_i + 1}{2} - 1$, so:
$$
\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i+(1-2)}{2}\right)\log\left(1 + \exp(w^Tx_i)\right)\\=
\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i+1}{2}-1\right)\log\left(1 + \exp(w^Tx_i)\right)
$$
Since $z_i = (y_i+1)/2$, $p(y_i\equiv1) = \left(1+\exp(-w^Tx_i)\right)^{-1}$, and $p(y_i\equiv-1) = \left(1 + \exp(w^Tx_i)\right)^{-1}$:
$$
\left(\frac{y_i+1}{2}\right)\log\left(1 + \exp(-w^Tx_i)\right)-
\left(\frac{y_i+1}{2}-1\right)\log\left(1 + \exp(w^Tx_i)\right)\\=
z_i\log\left(1/p(z_i\equiv1)\right)-
(z_i-1)\log\left(1/p(z_i\equiv0)\right)
$$
Next, a logarithm rule is that $\log(1/x) = -\log(x)$ for $x>0$.
$$
z_i\log\left(1/p(z_i\equiv1)\right)-
(z_i-1)\log\left(1/p(z_i\equiv0)\right)\\=
-z_i\log\left(p(z_i\equiv1)\right)+
(z_i-1)\log\left(p(z_i\equiv 0\right)
$$
Next, $p(z_i \equiv 0) = 1 - p(z_i \equiv 1)$, so:
$$
-z_i\log\left(p(z_i\equiv1)\right)+
(z_i-1)\log\left(p(z_i\equiv 0\right)\\=
-z_i\log\left(p(z_i\equiv1)\right)+
(z_i-1)\log\left(1-p(z_i\equiv1)\right)
$$
Next, factor out the minus sign.
$$
-z_i\log\left(p(z_i\equiv1)\right)+
(z_i-1)\log\left(1-p(z_i\equiv1)\right)\\=
{-\left(z_i\log\left(p(z_i\equiv1)\right)-
(z_i-1)\log\left(1-p(z_i\equiv1)\right)\right)}
$$
Finally, distribute the minus sign across the $z_i - 1$ on the right.
$$
{-\left(z_i\log\left(p(z_i\equiv1)\right)-
(z_i-1)\log\left(1-p(z_i\equiv1)\right)\right)}\\=
\color{blue}{-\left(z_i\log\left(p(z_i\equiv1)\right)+
(1-z_i)\log\left(1-p(z_i\equiv1)\right)\right)}
$$
With each summand equal in the logistic and log loss functions defined in the question, the two loss functions are equal.