Since convolution describes the operation of a linear time-invariant (LTI) system, the question is if the effect of an LTI system can be compensated by another LTI system. In the discrete-time domain you can use the $\mathcal{Z}$-transform to analyze LTI systems. If a signal $x(n)$ (with $\mathcal{Z}$-transform $X(z)$) is filtered by a system with impulse response $h(n)$ (with $\mathcal{Z}$-transform $H(z)$), then the $\mathcal{Z}$-transform of the resulting output signal $y(n)$ is given by
$$Y(z)=X(z)H(z)$$
Recovering $X(z)$ from $Y(z)$ can theoretically be done by
$$X(z)=\frac{Y(z)}{H(z)}$$
except at points $z$ where $H(z)=0$. The above equation corresponds to convolution with an impulse response $g(n)$ with $\mathcal{Z}$-transform
$$G(z)=\frac{1}{H(z)}\tag{1}$$
The question is if this system can be realized by a causal and stable filter. This is indeed the case if $H(z)$ is a minimum-phase system. Assuming rational transfer functions (which we always get when discrete-time systems consist of adders, multipliers, and delay elements), this means that $H(z)$ is a causal and stable system with all its zeros inside the unit circle of the complex plane (its poles are of course also inside the unit circle because of stability). Due the inversion in (1), the zeros of $H(z)$ are the poles of $G(z)$ and vice versa, i.e. if the zeros of $H(z)$ are inside the unit circle, then $G(z)$ is stable. One more thing that can be seen from (1) is that if $H(z)$ is an FIR system, then $G(z)$ is IIR. Nevertheless, in practice equalization if often done by FIR filters, which approximate the inverse filter in some optimal sense.
Let's look at a simple example. Assume an FIR system
$$h(n)=\delta(n)-a\delta(n-1),\quad |a|<1$$
Then we have
$$H(z)=1-az^{-1}$$
$H(z)$ is a minimum-phase system because it has one pole at $z=0$ and a zero at $z=a$, which is inside the unit circle if $|a|<1$. The inverse system is given by
$$G(z)=\frac{1}{1-az^{-1}}=\sum_{n=0}^{\infty}a^nz^{-n}$$
the impulse response of which is
$$g(n)=a^n,\quad n\ge 0$$
which is of course IIR.
So in this case the convolution
$$y(n)=(x*h)(n)=x(n)-ax(n-1)$$
can be perfectly compensated for by the convolution
$$x(n)=(y*g)(n)=\sum_{k=0}^{\infty}g(k)y(n-k)=\sum_{k=0}^{\infty}a^ky(n-k)$$