No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose $$ |H\cap S^{n_i}|/|S^{n_i}|\to 1 $$ for some subsequence $(n_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have $$ |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0. $$ Thus in particular $$ |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0. $$ But $$ S^{n_i-1} x \subset S^{n_i}, $$ so $$ S^{n_i-1} \subset S^{n_i} \cap S^{n_i} x^{-1}. $$ Also $S^{n_i} \subset S^{n_i-1} S$. Thus $$ |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-1}|/|S^{n_i}| \geq 1/|S| \not\to 0, $$ a contradiction.

But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$?

No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose $$ |H\cap S^{n_i}|/|S^{n_i}|\to 1 $$ for some subsequence $(n_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have $$ |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0. $$ Thus in particular $$ |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0. $$ But $$ S^{n_i-1} x \subset S^{n_i}, $$ so $$ S^{n_i-1} \subset S^{n_i} \cap S^{n_i} x^{-1}. $$ Also $S^{n_i} \subset S^{n_i-1} S$. Thus $$ |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-1}|/|S^{n_i}| \geq 1/|S| \not\to 0, $$ a contradiction.

But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$? However, we would need to have (1) lots of generators, (2) exponential growth. (If $|S^n|$ has subexponential growth then you have an invariant measure so you can't get bigger than upper density $1/2$.) Something in the free group $F_n$?

No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose $$ |H\cap S^{n_i}|/|S^{n_i}|\to 1 $$ for some subsequence $(n_i)$. Let $x\notin H$, and suppose $x\in S^m$. Since $H$ and $Hx$ are disjoint we have $$ |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0. $$ Thus in particular $$ |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0. $$ But $$ S^{n_i-m} x \subset S^{n_i}, $$ so $$ S^{n_i-m} \subset S^{n_i} \cap S^{n_i} x^{-1}. $$ Also $S^{n_i} \subset S^{n_i-m} S^m$. Thus $$ |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-m}|/|S^{n_i}| \geq 1/|S^m| \not\to 0, $$ a contradiction.

But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$?

No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose $$ |H\cap S^{n_i}|/|S^{n_i}|\to 1 $$ for some subsequence $(n_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have $$ |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0. $$ Thus in particular $$ |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0. $$ But $$ S^{n_i-1} x \subset S^{n_i}, $$ so $$ S^{n_i-1} \subset S^{n_i} \cap S^{n_i} x^{-1}. $$ Also $S^{n_i} \subset S^{n_i-1} S$. Thus $$ |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-1}|/|S^{n_i}| \geq 1/|S| \not\to 0, $$ a contradiction.

But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$?