[Edit: The question was later modified to say that $A$ is diagonal, which changes things quite a bit. This answer is for the general case. See Arturo's answer for the diagonal case.]
Typically, no, and Sivaram has already given an illustrative example. In your situation, the only exception is when $m=1$ and $b\neq 0$. The reason is that if $b$ is an $m$-by-$1$ vector and $m>1$, then $Ab$ never determines $A$ completely. One way to see this is to note that there exists an $m$-by-$m$ matrix $B$ such that $B$ is not the zero matrix, but $Bb=0$. Then $A+B\neq A$, but $(A+B)b=Ab$. Thus, whatever "$c/b$" might mean, it would have to be equally valid that it is equal to $A$ and to $A+B$, which is impossible.
In order for the equation $Ab=c$ to uniquely determine $A$, $b$ must be right invertible. If we generalized a bit to allow $b$ to have other sizes than $m$-by-$1$, say $m$-by-$k$, this will be possible when $b$ has at least as many columns as rows ($k\geq m$) and has maximal rank ($m$). This would correspond to $b$ representing a surjective linear transformation, and then $Ab$ determines $A$ because if you know $Ab$, then you know what $A$ does to everything in the range of $b$, namely everything. Algebraically speaking, if $d$ is a matrix such that $bd=I_m$, the $m$-by-$m$ identity matrix, then $A=AI_m=Abd=cd$. Thus, $cd$ plays the role of "$c/b$". Note however that unless $m=k$, $d$ is not uniquely determined by $b$. If $m=k$, then $d=b^{-1}$ is the inverse matrix of $b$, and $A=cb^{-1}$ looks more like the division you'd hope for. And again, if $k or for any other reason $b$ has rank less than $m$, $d$ does not exist.