No, the Beck-Chevalley condition does not hold for all pushout squares in a regular category, and not even if the category is exact, or a pretopos, or even a topos. In fact, here is a counterexample in $\rm Set$. Let $B = \{a,b\}$ and $C = \{\alpha,\beta\}$ and $A=\{0,1,2\}$, define $f:A\to B$ by $f(0)=f(1)=a$ and $f(2)=b$, and $g:A\to C$ by $g(0)=\alpha$ and $g(1)=g(2)=\beta$. Let $P$ be the pushout of $f$ and $g$ with coprojections $p:B\to P$ and $q:C\to P$. Then

$$q(\alpha) = q(g(0)) = p(f(0)) = p(a) = p(f(1)) = q(g(1)) = q(\beta) = q(g(2)) = p(f(2)) = p(b)$$

so $P$ is a one-element set. This, if we let $U=\{b\} \in S(B)$, then $p_!(U)$ is all of $P$, hence $q^* p_!(U)$ is all of $C$. (The notation $p_!$ for left adjoints of $p^*$ is, I think, more common than $p_*$ in this field, the latter being used more often for right adjoints.) But $f^*(U) = \{2\}$, so $g_! f^*(U) = \{\beta\} \neq q^* p_!(U)$.

There is something positive that can be said, however. Suppose $P = B \amalg_A C$ is a pushout of $f:A\to B$ and $g:A\to C$ such that the induced map $\Delta : A \to B\times_P C$ is a regular epimorphism. Then the pullback square defining $B\times_P C$, consisting of $h:B\times_P C\to B$ and $k:B\times_P C\to C$ with $p:B\to P$ and $q:C\to P$, does satisfy the Beck-Chevalley condition. Moreover, since $\Delta$ is a regular epimorphism, we have $\Delta_! \Delta^* = \rm Id$, and therefore

$$ g_! f^* = k_! \Delta_! \Delta^* h^* = k_! h^* = q^* p_! $$

so the original pushout square does satisfy the Beck-Chevalley condition.

The above "zig-zag" counterexample is simply the "minimal" situation in which $\Delta$ fails to be surjective. (One might call it "the minimal way to violate the hypotheses of the baby Blakers-Massey theorem".) Thus, it seems that $\Delta$ being a regular epi is probably the best possible hypothesis.