Voltage shift via op amp circuit
This can be very simple, and you only need one op-amp and four resistors. Given the values I've shown, assuming ideal components the solution is exact. As a hint on how to get to this solution, attempt to solve for two cases: the first for when the output is zero, and the second for when the output is equal to Vref.



simulate this circuit &ndash; Schematic created using CircuitLab
To get here:
If \$ V_{out} = 0 \$ , \$ R_3 \$ and \$ R_4 \$ form a voltage divider, \$ R_1 \$ and \$ R_2 \$ form a voltage divider, and \$ V_{in} = V_{off} \$ :
$$ V_n = \frac {V_{ref} R_4} {R_4 + R_3} $$
$$ V_p = \frac { V_{off} R_2 } {R_2 + R_1 } $$
$$ V_n = V_p \text{, so}$$
$$ \text{I. } 
V_{ref} R_4 \left(R_2 + R_1 \right) = V_{off} R_2 \left(R_4 + R_3\right)
$$
If \$ V_{out} = V_{ref} \$, there is no current through \$ V_n \$ so
$$ V_{out} = V_{ref} = V_{p} = V{n} $$
$$ \text{II. }
V_{ref} \left(R_2 + R_1 \right) =  \left(V_{ref} + V_{off} \right)
R_2  $$
Assign \$ R_1 / R_2 = a \$, \$ R_3 / R_4 = b \$ and you will find that \$ \text{I} \$ and \$ \text{II} \$ form a system of two equations, two unknowns. After a little algebra you will end up with
$$
\frac {R_1} {R_2} = \frac {R_4} {R_3} = \frac {V_{off}} {V_{ref}}
$$
There's probably an easier way to get this, but the above works from first principles.
You can co-scale R1,R2 and R3,R4 as desired. Since 3k and 8k are not E24 resistors, using (e.g.) 75k and 200k will still be exact but will be more accessible.