$g(x)=x^2-2$ works. (I'm taking $c$ to be the content, and $\ell$ to be the leading coefficient). Verification let to the reader, in case this is homework.
EDITed in the light of comments and edits to the question statement:
If non-constant $g$ has integer coefficients and leading coefficient $\pm1$, then $g(3/2)=\pm1$ is obviously impossible. However, $g(3/2)=\pm1$ has nothing to do with the ideal generated by $g$ and $2x-3$.
For example, if $g(x)=x^2-x-1$, then $g$ satisfies all your conditions: it is irreducible, has integer coefficients, has leading coefficient and constant term $\pm1$, and with $2x-3$ generates the ring ${\bf Z}[x]$, as is evident from $(-4)(x^2-x-1)+(2x+1)(2x-3)=1$
I don't know whether there is a polynomial that satisfies your conditions simultaneously for $2x-3$ and $5x+7$, but if I find one, I'll let you know.
Further EDIT: You say I can change the polynomials. If I change the $5x+7$ to $5x-8$ then $x^2-x-1$ will solve your problem.
Even more EDIT: I'm now confident (but not 100% certain, since I haven't carried out all the calculations) that there is a polynomial $g$ of degree 6 with integer coefficients, leading coefficient 1, constant term $-1$, irreducible over the rationals, such that $1$ is in both the ideals $(g,5x+7)$ and $(g,2x-3)$.
The condition on the ideals will be satisfied if $5^6g(-7/5)=-1$ and $2^6g(3/2)=-1$. Let $g(x)=x^6+ax^5+bx^4+cx^3+dx^2+ex-1$ Then we get the two equations in $5$ unknowns, $7^6-7^55a+7^45^2b-7^35^3c+7^25^4d-(7)5^5e-5^6=-1$ and $3^6+3^52a+3^42^2b+3^32^3c+3^22^4d+(3)2^5e-2^6=-1$ Move the constant terms to the right side of the equations, divide the 1st one by $-35$ (note that $7^6-5^6+1$ is a multiple of $35$) and the second one by $6$ ($3^6-2^6+1$ is a multiple of $6$), and you have two linear equations in $5$ unknowns, with no modular obstacle to a solution.
Now I wave my hands a little and say there must be infinitely many integer solutions to this pair of equations, including infinitely many with $g$ irreducible. In any event, it should not be hard to find one such solution.
Please get back to me if there are any questions about this.