Indeed, there is at most one root on $x\in[0,\infty)$ -- actually, for each real $a$. Consider the equivalent equation
\begin{equation}
r(x):=\frac{f(x)}{g(x)}=\frac{\sqrt\pi}b \tag{1}
\end{equation}
for real $x>z_a:=0\vee(-a)$,
where $0<b<1$ and
\begin{equation}
f(x):=\frac{e^{-x^2}}{x+a},\quad g(x):=1-\text{erf}(x).
\end{equation}
Consider the "derivative ratio"
\begin{equation}
\rho(x):=\frac{f'(x)}{g'(x)}=\frac{\sqrt{\pi } \left(1+2 a x+2 x^2\right)}{2 (x+a)^2}.
\end{equation}
Then
\begin{equation}
\rho'(x)=\frac{\sqrt{\pi } \left(a^2+a x-1\right)}{(x+a)^3},
\end{equation}
so that, for real $x>z_a$, we have
$$\rho'(x)>0 \iff a>1\text{ or } \big(0 < a \le 1\ \&\ x > x_a:=(1 - a^2)/a\big).$$

Now we are ready to use l'Hospital-type rules for monotonicity.

*Case 1:* $a\le0$, so that $z_a=-a$. Then $\rho$ is (strictly) decreasing (on the entire interval $(z_a,\infty)$),
whence, by Proposition 4.1 in the mentioned paper, $r$ is decreasing, and so, equation (1) has at most one root (in $(z_a,\infty)$). In fact, in this case there is exactly one root, since $r(z_a+)=\infty$ and $r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$.

*Case 2:* $0<a\le1$, so that $z_a=0$. Then $\rho$ is decreasing on $(z_a,x_a)$ and increasing on $(x_a,\infty)$, whence, by Proposition 4.3 in the same paper, $r$ is decreasing on $(z_a,c_a)$ and increasing on $(c_a,\infty)$, for some real $c_a\ge z_a$. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$ on $[c_a,\infty)$. Hence, equation (1) has at most one root in $(z_a,\infty)$, and that possible root must be in $(z_a,c_a)$.
In fact, if $0<a\le b/\sqrt\pi$, then there is exactly one root in $[0,\infty)$, since $r(0)=\frac1a\ge\frac{\sqrt\pi}b$ and $r(c_a)<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$.
Accordingly, if $b/\sqrt\pi<a\le1$, then there is no root.

*Case 3:* $a>1$. Then $\rho$ is increasing (on the entire interval $(z_a,\infty)$),
whence, by the mentioned Proposition 4.1, $r$ is increasing. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. Thus, equation (1) has no roots (in $(z_a,\infty)$) in this case.