A sufficient condition that $\tau_a(n)$ grows at most polynomially in $n$ is that $\prod_{r=1}^\infty (1-q^{a_r})$ be a expressible as a finite product $S_1 \ldots S_m$ of series with bounded coefficients. (Let $S_i=\sum_{j=1}^\infty c_{i,j} q^j$, and let $B_j$ be the bound for the coefficients of $S_j$. $\tau_a(n)=\sum_{t_1+\cdots +t_m=n} c_{1,t_1} \cdots c_{1,t_m}$, hence $|\tau_a(n)| \le B_1 \ldots B_m n^m$.)

Thus, it follows from a 24-fold application of the pentagonal number theorem, $$\prod_{n=1}^\infty (1-x^n)=\sum_{k=-\infty}^\infty(-1)^kx^{k(3k-1)/2},$$ that $\tau(n)$ grows polynomially. (This argument presumably goes back to Ramanujan.)

The condition of having the density of even terms bounded below is not sufficient to have $\tau_a(n)$ be bounded by a polynomial; $a_n=4n-2$ is a counterexample by a similar argument to my previous answer (replace $q$ with $iq$). We can get rid of counterexamples in that vein by requiring that the sequence $a_n$ has an infinite subsequence $b_n$ such that $\tau_b$ is bounded, which leads me to wonder what bearing the boundedness of $\tau_b(n)$ for a subsequence $b$ of $a$ has on the polynomial boundedness of $\tau_a(n)$, i.e., is it conceivably necessary or sufficient?

A sufficient condition that $\tau_a(n)$ grows at most polynomially in $n$ is that $\prod_{r=1}^\infty (1-q^{a_r})$ be a expressible as a finite product $S_1 \ldots S_m$ of series with bounded coefficients. (Let $S_i=\sum_{j=1}^\infty c_{i,j} q^j$, and let $B_j$ be the bound for the coefficients of $S_j$. $\tau_a(n)=\sum_{t_1+\cdots +t_m=n} c_{1,t_1} \cdots c_{1,t_m}$, hence $|\tau_a(n)| \le B_1 \ldots B_m n^m$.)

Thus, it follows from a 24-fold application of the pentagonal number theorem, $$\prod_{n=1}^\infty (1-x^n)=\sum_{k=-\infty}^\infty(-1)^kx^{k(3k-1)/2},$$ that $\tau(n)$ grows polynomially. (This argument presumably goes back to Ramanujan.)

The condition of having the density of even terms bounded below is not sufficient to have $\tau_a(n)$ be bounded by a polynomial; $a_n=4n-2$ is a counterexample by a similar argument to my previous answer (replace $q$ with $iq$). We can get rid of counterexamples in that vein by requiring that the sequence $a_n$ has an infinite subsequence $b_n$ such that $\tau_b$ is bounded, which leads me to wonder what bearing the boundedness of $\tau_b(n)$ for an infinite subsequence $b$ of $a$ has on the polynomial boundedness of $\tau_a(n)$, i.e., is it conceivably necessary or sufficient?