I'm going to assume $R$ contains $1$.
Also, my proof will only show that $RbR \subseteq aR \cap Ra$.
Most definitions of total divisors I've seen go like the following: An element $a$ in a ring $R$ is a total divisor of $b$ when $RbR \subseteq aR \cap Ra$. Whether this implies $RbR = aR \cap Ra$ in a ring that is both a left and right principal ideal domain with unity, I'm unsure of. Maybe you could provide information on where you found this problem?
Since $RbR$ is an ideal, it is also a right ideal, and because $R$ is a right principal ideal domain, we get $RbR = Rx$ for some $x \in R$. Again, because $R$ is a right PID, and sums of right ideals are right ideals, we have $Ra + Rx = Rd$ for some $d \in R$. Thus, $d = r_1a +r_2x$, with $r_1, r_2 \in R$. Further, $x = ar$ for some $r \in R$, because, $RbR \subseteq aR$ and $1 \in R$.
Now, $dr = r_1ar +r_2xr = r_1ar + r_2r'x$ (for some $r' \in R$, because $xR = Rx$) $= r_1ar + r_2r'ar = (r_1a + r_2r'a)r$ $\implies d = r_1a + r_2r'a$ when $r \neq 0$, that is when $a \neq 0$ (should we have $a = 0$ the result follows trivially, so let's assume $a \neq 0$).
That is, $d \in Ra$, so $Rd \subseteq Ra$, and $Rd = Ra$. From our previous equation $Ra + Rx = Rd$, we see that $Rx \subseteq Ra$. But $RbR = Rx = xR$, so $RbR \subseteq Ra$. We have, adding the hypothesis, that $RbR \subseteq aR$ and $RbR \subseteq Ra$, thus - $RbR \subseteq aR \cap Ra$.