It's not true that it works for $Z$ small enough. Consider the $2 \times 2$ case $$ Z = \pmatrix{t & 2t\cr 2t & 4t\cr} $$ which is positive semidefinite for $t \ge 0$. Then $$\det(X) = \log(1+t)\log(1+4t) - \log(1+2t)^2 $$ which appears to be negative for all $t > 0$, and certainly is negative for small $t > 0$: its Maclaurin series is $ \det(X) = -2 t^3 + O(t^4)$.

What is true is that if $Z$ is positive definite, $\log(1+tZ)$ will be positive definite for sufficiently small $t > 0$. This can be obtained using the Maclaurin series for $\log(1+z)$.

It's not true that it works for $Z$ small enough. Consider the $2 \times 2$ case $$ Z = \pmatrix{t & 2t\cr 2t & 4t\cr} $$ which is positive semidefinite for $t \ge 0$. Then $$\det(X) = \log(1+t)\log(1+4t) - \log(1+2t)^2 $$ which appears to be negative for all $t > 0$, and certainly is negative for small $t > 0$: its Maclaurin series is $ \det(X) = -2 t^3 + O(t^4)$.

What is true is that if $Z$ is positive definite, $\log(1+tZ)$ will be positive definite for sufficiently small $t > 0$. This can be obtained using the Maclaurin series for $\log(1+z)$.

EDIT:

Let $g(z) = z - \log(1+z)$, and write $X = Z - G$, where $G_{ij} = g(Z_{ij})$. Note that $g$ is increasing on $[0,1]$. If all $|Z_{ij}| \le b$, then $|W_{ij}| \le g(b)$. If your matrices are $N \times N$, for any vector $x$ we have by Cauchy-Schwarz $x^T G x \le N g(b) \|x\|^2$. Thus to have $X \succeq 0$ it suffices for $Z$ to be positive definite with least eigenvalue $\lambda$ and elementwise bound $b$ where $$ \lambda \ge N g(b) $$