Does there exist a Hamel basis $\mathcal B$ for $\mathbb R$ over $\mathbb Q$ such that $a,b \in \mathcal B \implies \dfrac ab \in \mathcal B$?
The solution follows closely from the standard argument in group theory.
Suppose that a Hamel basis $\mathcal{B}$ satisfies the property $$ a, b\in \mathcal{B} \Longrightarrow \frac ab\in \mathcal{B}. $$ From this, we have $$ a\in\mathcal{B} \Longrightarrow \frac aa =1\in \mathcal{B}. $$ Then $$ b\in\mathcal{B} \Longrightarrow \frac 1b \in \mathcal{B}. $$ Now, $$ a,b \in \mathcal{B} \Longrightarrow a, \frac 1b \in \mathcal{B} \Longrightarrow \frac a{1/b} = ab \in \mathcal{B}. $$ Since $\mathcal{B}$ contains some element $a\neq 0, 1$, a contradiction comes from the same argument as in Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$.
For the new condition by Noah Schweber
If $1\in\mathcal{B}$ then it can be proven the same way as above. So, we suppose that $1\notin \mathcal{B}$. Then we have $1$ as a $\mathbb{Q}$-linear combination of elements in $\mathcal{B}$, say $$ 1=\sum_{i=1}^{k} \epsilon_i x_i $$ where $x_i\in\mathcal{B}$, $\epsilon_i\in\mathbb{Q}$.
Choose $y_1\in \mathcal{B}$. We modify the argument in Jonathan Golan's answer to Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$.
Let $\mathcal{B}_s= \mathcal{B} - \mathbb{Q}(y_1, x_1, \ldots, x_k)$ be a selected basis which are not rational in $y_1$, $x_i$'s. Let $\alpha: \mathbb{R} \rightarrow \mathbb{Q}$ defined by $$ r=\sum_{x\in\mathcal{B}} r_x x \mapsto \alpha(r)=\sum_{x\in\mathcal{B}_s} r_x. $$ Then for any $x\in \mathcal{B}_s$, we have $\frac x{y_1}\in\mathcal{B}_s$. Let $a=\frac1{y_1}$. For any $r\in \mathbb{R}$, we have $\alpha(r) = \alpha(ar)$.
For any $x\in \mathcal{B}_s$, put $ r=x(a-1)^{-1}. $
Then $$ 1=\alpha(x) = \alpha( r(a-1) ) = \alpha( ar - r) = \alpha(ar)-\alpha(r) = 0. $$ This is a contradiction.
This is not an answer. It is an attempt at Noah Schweber's new condition. See his comment under i707107's solution.
Let $F$ be a field extension of a field $K$ with $n:=[F:K]\geq 4$. Suppose that $B$ is a Hamel basis of $F$ over $K$ with the divisibility property, namely, for all $a,b\in B$ with $a\neq b$, $\dfrac{a}{b}\in B$. Then, it holds that $ab\in B$ for all $a,b\in B$ with $ab\neq 1$.
Fix $a,b\in B$ with $ab\neq 1$. Then, choose $c\in B\setminus\{a\}$ and $d\in B\setminus\left\{c,bc,\dfrac{c}{a}\right\}$. Hence, $\dfrac{c}{a}\in B$ and $\dfrac{ad}{c}=\dfrac{d}{c/a}\in B$. Furthermore, $\dfrac{d}{c}\in B$ and $\dfrac{d}{bc}=\dfrac{d/c}{b}\in B$. Since $ab\neq 1$, we have $\dfrac{ad}{c}\neq \dfrac{d}{bc}$. Consequently, $$ab=\frac{ad/c}{d/bc}\in B\,.$$
Involutive Hamel Bases
A subset $S$ of $F\setminus\{0\}$ is said to be involutive if $S$ is invariant under inversion (that is, $\dfrac{1}{s}\in S$ for all $s\in S$). Now, if we can show that every Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ is not involutive, then it follows that a Hamel basis with the divisibility property does not exist. I don't think that an involutive Hamel basis for the extension $\mathbb{R}>\mathbb{Q}$ exists, but I have no idea about a proof. More generally, I would like to know whether there exist a field $K$ and an infinite field extension $F$ of $K$ with a Hamel basis $B$ over $K$ such that $B$ is involutive. As in i707107's solution, involutivity of $B$ is not needed if $[F:K]>|\bar{K}|$, where $\bar{K}$ is the algebraic closure of $K$, but it is an interesting question nonetheless.
If $n=[F:K]$ is finite and odd, then $$B=\left\{\bar{x}^{-\left\lfloor\frac{n}{2}\right\rfloor},\bar{x}^{-\big(\left\lfloor\frac{n}{2}\right\rfloor-1\big)},\ldots,\bar{x}^{+\big(\left\lfloor\frac{n}{2}\right\rfloor-1\big)},\bar{x}^{+\left\lfloor\frac{n}{2}\right\rfloor}\right\}$$ is an involutive basis of $F$ over $K$, where $F=K[x]\big/\big(f(x)\big)$ for some irreducible polynomial $f(x)\in K[x]$. If $n$ is even, then there exists an irreducible polynomial $f(x)$ of degree $n$ in $K[x]$ such that the coefficient of the term $x^{n/2}$ is nonzero and that $F=K[x]\big/\big(f(x)\big)$, making $$B=\left\{\bar{x}^{-1},\bar{x}^{-2},\ldots,\bar{x}^{-(n/2)}\right\}\cup\left\{\bar{x}^{+1},\bar{x}^{+2},\ldots,\bar{x}^{+(n/2)}\right\}$$ an involutive Hamel basis of $F$ over $K$. (For example, with $F=\mathbb{C}$ and $K=\mathbb{R}$, we can take $f(x)=(x+1)^2+1$.)
Existence of $B$ with the Divisibility Property
For $n=2$, take $K:=\mathbb{F}_2$ and $F:=\mathbb{F}_4=K[x]\big/\left(x^2+x+1\right)$. Then, $B:=\left\{\bar{x},\bar{x}+1\right\}$ satisfies the condition, where $\bar{x}$ is the image of $x$ under the canonical projection $K[x]\to K[x]\big/\left(x^2+x+1\right)$. In fact, if such a Hamel basis $B$ exists for the case $n=2$, then $x^2+x+1$ is an irreducible polynomial in $K[x]$ and $F=K[x]\big/\left(x^2+x+1\right)$, in which case $B=\left\{\bar{x},-\bar{x}-1\right\}$.
For $n=3$, it turns out that $B$ cannot exist. Suppose contrary that $B=\{a,b,c\}$ exists. Then, we can easily see that $B$ does not contain $1$ (or the extension would be of index at most $2$). Hence, $\dfrac{a}{b}\neq a$ and $\dfrac{a}{c}\neq a$. If $\dfrac{a}{b}=c$, then $c\notin K$, whence $\dfrac{b}{a}=\dfrac{1}{c}\neq c$ so that $\dfrac{b}{a}=a$ is the only possibility. Furthermore, we also have $\dfrac{a}{c}=b$, which leads to $b\notin K$ and $\dfrac{c}{a}=a$. Therefore, $b=a^2=c$, a contradiction. Hence, $\dfrac{a}{b}=b$ and $\dfrac{a}{c}=c$. However, this gives $b^2=a=c^2$, or $b=\pm c$, which is absurd.
As i707107 shows, $B$ does not exist if $[F:K]>|\bar{K}|$. Replace $\mathbb{Q}$ in i707107's solution by $K$ and $\mathbb{R}$ by $F$ to get a proof of this claim. We are left to deal with the case where $[F:K]>3$ is finite and the case when $[F:K]$ is infinite but $[F:K]\leq|\bar{K}|$.
Case 1: The index $n=[F:K]$ is an odd integer greater than $3$. Suppose $B$ exists. It is evident that $1\notin B$ and $B$ must be involutive. Since $n=|B|$ is odd, $B$ has an involutive element $u\in B$ with $u=\dfrac{1}{u}$. Because $1\notin B$, we have $u=-1$ (whence the characteristic of $K$ cannot be $2$). Thus, for any $a\in B\setminus\{u\}$, we have $\dfrac{u}{a}=-\dfrac{1}{a}$ must lie in $B$. This contradicts the fact that $B$ is involutive (which means $\dfrac{1}{a}$ is in $B$).
Case 2: The index $n=[F:K]$ is an even integer greater than $2$. If $B$ exists, then $B$ can be partitioned into $\left\{t^{+j},t^{-j}\right\}$ for some $t\in F$ and $j=1,2,\ldots,\frac{n}{2}$. Ergo, $t^p=1$ for some integer $p>0$. If $p$ is not prime, then we can see that $B$ is not $K$-linearly independent, which is absurd. Hence, we see that $[F:K]=n=p-1$ for some odd prime natural number $p$ and $$B=\left\{t,t^2,\ldots,t^{p-1}\right\}=\bigcup\limits_{j=1}^{\frac{p-1}{2}}\,\left\{t^{+j},t^{-j}\right\}$$ for some primitive $p$-th root of unity $t\in F\setminus K$. This is possible only if $\text{char}(K)\neq p$. In summary, for the case $[F:K]$ is even, $B$ exists if and only if $F$ is a cyclotomic field extension over $K$ generated by a primitive $p$-root of unity with $p$ being an odd prime.
Case 3: The index $n=[F:K]$ is infinite and $n\leq|\bar{K}|$. If $F$ is algebraic over $K$, then we can follow the two former cases that such a Hamel basis $B$ would have to contain only primitive $p$-roots of unity for odd prime natural numbers $p$, but this is absurd as products of most pairs in $B$ are also in $B$. Hence, $B$ does not exist if $F$ is algebraic over $K$. The subcase where $F$ is not purely algebraic over $K$ seems to be very difficult.