The polynomial hierarchy would collapse to the fourth level (indeed, the third, see below).

Proof. First, we get that $\oplus P\subseteq \Sigma_2^P$. This is because a $\Sigma_2^P$ machine can use its $\exists$ quantifier to guess a polynomial number of assignments and then, using its $\forall$ quantifier, verify that this list contains all of the formula's satisfying assignments. By Valiant-Vazirani we have $PH\subseteq BPP^{\oplus P}$. We have $BPP\subseteq \Sigma_2^P$, so we get $BPP^{\oplus P}\subseteq {\Sigma_2^P}^{\oplus P}\subseteq {\Sigma_2^P}^{\Sigma_2^P}=\Sigma_4^P=PH$. $\square$

@Emil Jerábek gives a stronger argument; the polynomial hierarchy collapses to the third level. This is because counting assignments can be done in $P^{NP}$, which is better than the $\Sigma_2^P$ method described above.

Proof. We get $\oplus P\subseteq P^{NP}$, because a $P^{NP}$ algorithm can count a formula's satisfying assignments, if it is polynomially-bounded. (algorithm below). Consequently, we get $PH\subseteq BPP^{\oplus P}\subseteq BPP^{P^{NP}}\subseteq \left(\Sigma_2^P\right)^{P^{NP}}=\left(\Sigma_2^P\right)^{NP}=\Sigma_3^P=PH$.

Algorithm [For counting satisfying assignments in $P^{NP}$].$\quad$ Given a Boolean formula $\psi$, query the oracle. If the formula is not satisfiable, return. Otherwise, find a satisfying assignment $x$ using $n$ more queries to the oracle. Remove the assignment, i.e., construct a new formula $\psi^\prime:=\psi\wedge \neg x$. Repeat this procedure until the oracle returns ``not satisfiable''. This happens exactly after all the satisfying assignments have been "removed". This procedure therefore runs in time polynomial in the number of satisfying assignments, which, by assumption, is polynomial.

The inclusion $\oplus P\subseteq P^{NP}$ is obtained by simply returning ``yes'' at the end of the computation if and only if the number of satisfying assignments is discovered to be odd. $\square$

Perhaps there are other consequences; I hope others can give some.

The polynomial hierarchy would collapse to the fourth level (indeed, the third, see below).

Proof. First, we get that $\oplus P\subseteq \Sigma_2^P$. This is because a $\Sigma_2^P$ machine can use its $\exists$ quantifier to guess a polynomial number of assignments and then, using its $\forall$ quantifier, verify that this list contains all of the formula's satisfying assignments. By Valiant-Vazirani we have $PH\subseteq BPP^{\oplus P}$. We have $BPP\subseteq \Sigma_2^P$, so we get $BPP^{\oplus P}\subseteq {\Sigma_2^P}^{\oplus P}\subseteq {\Sigma_2^P}^{\Sigma_2^P}=\Sigma_4^P=PH$. $\square$

@Emil Jerábek gives a stronger argument; the polynomial hierarchy collapses to the third level. This is because counting assignments can be done in $P^{NP}$, which is better than the $\Sigma_2^P$ method described above.

Proof. We get $\oplus P\subseteq P^{NP}$, because a $P^{NP}$ algorithm can count a formula's satisfying assignments, if it is polynomially-bounded. (algorithm below). Consequently, we get $PH\subseteq BPP^{\oplus P}\subseteq BPP^{P^{NP}}=BPP^{NP}\subseteq \left(\Sigma_2^P\right)^{NP}=\Sigma_3^P=PH$.

Algorithm [For counting satisfying assignments in $P^{NP}$].$\quad$ Given a Boolean formula $\psi$, query the oracle. If the formula is not satisfiable, return. Otherwise, find a satisfying assignment $x$ using $n$ more queries to the oracle. Remove the assignment, i.e., construct a new formula $\psi^\prime:=\psi\wedge \neg x$. Repeat this procedure until the oracle returns ``not satisfiable''. This happens exactly after all the satisfying assignments have been "removed". This procedure therefore runs in time polynomial in the number of satisfying assignments, which, by assumption, is polynomial.

The inclusion $\oplus P\subseteq P^{NP}$ is obtained by simply returning ``yes'' at the end of the computation if and only if the number of satisfying assignments is discovered to be odd. $\square$

Perhaps there are other consequences; I hope others can give some.

The polynomial hierarchy would collapse to the third level (possibly lower).

Proof. First, we get that $\oplus P\subseteq \Sigma_2^P$. This is because a $\Sigma_2^P$ machine can use its $\exists$ quantifier to guess a polynomial number of assignments and then, using its $\forall$ quantifier, verify that this list contains all of the formula's satisfying assignments. By Valiant-Vazirani we have $PH\subseteq RP^{\oplus P}$, which in our situation yields $RP^{\oplus P}\subseteq NP^{\oplus P}\subseteq NP^{\Sigma_2^P}=\Sigma_3^P=PH$. $\square$

Perhaps there are other consequences; I hope others can give some.

The polynomial hierarchy would collapse to the fourth level (indeed, the third, see below).

Proof. First, we get that $\oplus P\subseteq \Sigma_2^P$. This is because a $\Sigma_2^P$ machine can use its $\exists$ quantifier to guess a polynomial number of assignments and then, using its $\forall$ quantifier, verify that this list contains all of the formula's satisfying assignments. By Valiant-Vazirani we have $PH\subseteq BPP^{\oplus P}$. We have $BPP\subseteq \Sigma_2^P$, so we get $BPP^{\oplus P}\subseteq {\Sigma_2^P}^{\oplus P}\subseteq {\Sigma_2^P}^{\Sigma_2^P}=\Sigma_4^P=PH$. $\square$

@Emil Jerábek gives a stronger argument; the polynomial hierarchy collapses to the third level. This is because counting assignments can be done in $P^{NP}$, which is better than the $\Sigma_2^P$ method described above.

Proof. We get $\oplus P\subseteq P^{NP}$, because a $P^{NP}$ algorithm can count a formula's satisfying assignments, if it is polynomially-bounded. (algorithm below). Consequently, we get $PH\subseteq BPP^{\oplus P}\subseteq BPP^{P^{NP}}\subseteq \left(\Sigma_2^P\right)^{P^{NP}}=\left(\Sigma_2^P\right)^{NP}=\Sigma_3^P=PH$.

Algorithm [For counting satisfying assignments in $P^{NP}$].$\quad$ Given a Boolean formula $\psi$, query the oracle. If the formula is not satisfiable, return. Otherwise, find a satisfying assignment $x$ using $n$ more queries to the oracle. Remove the assignment, i.e., construct a new formula $\psi^\prime:=\psi\wedge \neg x$. Repeat this procedure until the oracle returns ``not satisfiable''. This happens exactly after all the satisfying assignments have been "removed". This procedure therefore runs in time polynomial in the number of satisfying assignments, which, by assumption, is polynomial.

The inclusion $\oplus P\subseteq P^{NP}$ is obtained by simply returning ``yes'' at the end of the computation if and only if the number of satisfying assignments is discovered to be odd. $\square$

Perhaps there are other consequences; I hope others can give some.