There is no single continuous $f:\ {\mathbb R}^2\to{\mathbb R}$ that produces a nonempty bounded curve for all $a>0$.
Proof. Let $g(x,y):=y\ f(x,y)$. Then $\inf g(x,y)=-\infty$, or else the solution set $S_a:=\{(x,y)\ |\ a + y\ f(x,y)=0\}$ would be empty for large enough $a$. Therefore for arbitrary $R>a$ one can find a point $(\xi,\eta)\in{\mathbb R}^2$ with $\xi^2+\eta^2>R^2$ and $g(\xi,\eta)<-R$. Connect the point $(R,0)$ with $(\xi,\eta)$ by a curve $\gamma:\ t\to z(t)$ $\ (0\leq t\leq 1)$ such that $|z(t)|\geq R$ for all $t\in[0,1]$. As $g(R,0)=0$ and $g(\xi,\eta)<-R$ there has to be a $\tau\in[0,1]$ with $g\bigl(z(\tau)\bigr)=-a$. It follows that $z(\tau)\in S_a$ and at the same time $|z(\tau)|\geq R$. As $R>a$ was arbitrary, $S_a$ is unbounded.$\quad {}_\square$
When $f$ is allowed to depend on $a$ (or $a$ is bounded away from $\infty$) then one can construct examples: For a fixed $b>0$ consider the function $f(x,y):={2b\over 1+x^2+y^2}\ .$ I claim that the solution set $S_a$ is a circle for any $a$ with $0.
Proof. Since $1+x^2+y^2>0$ for all $(x,y)$ the equation $a+y f(x,y)=0$ is equivalent with $(1+x^2+y^2) + 2y{b\over a}=0$ or $x^2 +\Bigl(y+{b\over a}\Bigr)^2={b^2-a^2\over a^2}\ .$ When $0 this is the equation of a circle in the $(x,y)$-plane. $\quad{}_\square$