If the entries of a positive semidefinite matrix shrink individually, will the operator norm always decrease?

It is not always true.

The following matrix is positive semidefinite with norm $3$: $$ P := \left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & -1\\ 1 & -1 & 2\\ \end{array}\right) $$ Use $B$ to poke out the $-1$'s and you get $$ P \circ B = \left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & 0\\ 1 & 0 & 2\\ \end{array}\right), $$ which is positive semidefinite with norm $2 + \sqrt{2} > 3$.

I think this is true. Here's an attempt that looks potentially fruitful:

Use the property given here. That is, note that $$ \DeclareMathOperator{\tr}{tr} \| P \circ B\| = \sup_{\|x\| = \|y\| = 1} x^*(P \circ B)y = \sup_{\|x\| = \|y\| = 1} \tr (D_x P D_y B^T) = \\ \sup_{\|x\| = \|y\| = 1} \langle D_x P D_y,B \rangle \leq \sup_{\|x\| = \|y\| = 1} \left(\sum_{i,j}|(D_x P D_y)_{ij}|\right)\max_{i,j}|B_{ij}| $$ here, $\langle \cdot , \cdot \rangle$ is an entry-wise dot-product, and $D_x = \operatorname{diag}(x_1,\dots,x_n)$. From here, maybe you can use the fact that $P$ can be written as a convex combination of PSD rank $1$ matrices. Perhaps it's useful to note that $D_x vv^T D_y = (x \circ v) (y \circ v)^T$.