Does $d(x,y) = \log(1 + | x - y |)$ define a metric on $\mathbb{R}$?

Yes, your proof that $d(x,y)=0$ implies $x=y$ makes sense. All you need is the fact that the $\log(s)$ function on $s > 0$ has only one zero, and that zero is $s=1$. Therefore, if $\log(s) = 0$, then $s$ can only be equal to $1$, hence if $1 = s = 1 + |x-y|$, then $|x-y|=0$, which implies $x=0$.

As for the triangle inequality, ask, for a general function $f(s), s > 0$, how the condition $$ f(a+b) \leq f(a) + f(b) $$ relates to the concavity of that function. Examine a few examples: e.g., the convex function $f(s) = s^2$, the concave function $f(s)=s^{1/3}$, the linear function $f(s) = 7s$.

And note that, for nonnegative $a, b$, $$ \log(1 + a + b) \leq \log(1 + a + 1 + b). $$

The $d\left(x,y\right)=0$ proof is correct. For the triangle inequality, I would start out with the right hand side. Notice that

\begin{align*} \log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)&=\log\left(\left(1+\left|x-z\right|\right)\left(1+\left|z-y\right|\right)\right)\\ &=\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right). \end{align*}

Recall that for all $a,b>0$, $\log\left(a\right)\ge\log\left(b\right)\iff a\ge b$. By the triangle inequality, $\left|x-y\right|\le\left|x-z\right|+\left|z-y\right|$, and so \begin{align*} 1+\left|x-y\right|&\le1+\left|x-z\right|+\left|z-y\right|\\ &\le1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|. \end{align*}

Thus, \begin{align*} d\left(x,y\right)=\log\left(1+\left|x-y\right|\right)&\le\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right)\\ &=\log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)\\ &=d\left(x,z\right)+d\left(z,y\right). \end{align*}