Does this help?
var('p s t')
def transf(expr):
w0 = SR.wild(0); w1 = SR.wild(1);
expr_t = expr.subs({p^s:t})
expr_t = expr_t.subs({p^(w0*s):t^w0})
expr_t = expr_t.subs({p^(s+w1):t*p^w1})
expr_t = expr_t.subs({p^(w0*s+w1):t^w0*p^w1})
return expr_t
# expression 1
expr1 = (p^(s + 3) - 1)*(p - 1)*p^(2*s)/((p^(s + 1) - 1)*(p^(2*s + 3) - 1))
# expression 2
expr2 = p^s / (p^(s+1) - 1)
for expr in [expr1, expr2]:
pretty_print(expr, LatexExpr(r"\mapsto"), transf(expr))
produces:
$$
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{{\left(p - 1\right)} p^{2 \, s} {\left(p^{s + 3} - 1\right)}}{{\left(p^{2 \, s + 3} - 1\right)} {\left(p^{s + 1} - 1\right)}} \mapsto \frac{{\left(p^{3} t - 1\right)} {\left(p - 1\right)} t^{2}}{{\left(p^{3} t^{2} - 1\right)} {\left(p t - 1\right)}}
$$
$$
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{p^{s}}{p^{s + 1} - 1} \mapsto \frac{t}{p t - 1}
$$
My limited knowledge of using wild cards is that you have to be very careful because it does not identify "units", and "zero"; see the last substitution would be fine mathematically to catch all instances (if wildcard meant anything) but since the expression is syntactically different from the others, we need the previous ones.
