This circuit might suit your needs.

^{simulate this circuit – Schematic created using CircuitLab}

Here is a sweep of output voltage vs input voltage.

The same sweep of output voltage vs input voltage can be accomplished with a single op-amp as well. However, the circuit above has the advantage that from the point of view of the input, the circuit appears as a 10 k\$\Omega\$ (or whatever value you choose) resistor to ground. With the single op-amp version, the circuit will appear as a resistor connected to the offset voltage, at least in the simplest configuration. Multiple op-amps in a single package are common.

Your circuit won't do what you want. It has no negative feedback, so the output will be either +5V or -5V, almost never in between. If you use an op-amp in this role, you must control gain with negative feedback, to obtain a controlled, analogue output.

If you don't mind losing some precision, the simplest way to convert a -2.5V to +2.5V signal to an exclusively positive range is with a simple resistor potential divider:

^{simulate this circuit – Schematic created using CircuitLab}

This produces a signal between +92mV and +3.36V, corresponding to an input ranging from -2.5V and +2.5V. That's not the full 5V range, but it might suffice, without requiring an op-amp or a negative supply. You can calibrate (scale and offset) the value in software.

If you require a larger range, then you must have voltage gain, which mandates an amplifier. Be aware that no op-amp can produce an output that extends all the way to its supply potentials. For example, if you use a TLC2272 op-amp with 0V and +5V supplies, you cannot expect an output outside of the range +20mV to +4.98V or so. My suggestions below will output a potential in the range +50mV to +4.95V, corresponding to an input of -2.5V to +2.5V. You'll need to account for this slightly-less-than-unity gain in software. The op-amp you employ should have rail-to-rail output capability, such as the TLC2272.

This circuit requires no negative supply:

^{simulate this circuit}

If you have access to a negative supply, -5V, then the resistor count can be reduced:

^{simulate this circuit}

It seems simpler, but you'll probably have to use compound resistors in series or parallel to produce those unusual values, so there may be no real benefit. The above circuit also inverts, so the output goes from +4.95V down to +50mV as the input rises from -2.5V to +2.5V, which you'll have to compensate for in software:

The best output range I could obtain using only standard E24 resistor values is this next one:

^{simulate this circuit}

This produces an output between +4.8V and +0.29V, as the input goes from -2.5V to +2.5V.

A differential amplifier with a gain of 1 and 2.5V as offset. Theoretically no need for a negative supply for the opamp, but if it has to go all the way to 0 it is properly needed.

A need of circuit “philosophy”

The only thing that remains after these good specific circuit solutions is to summarize them by revealing the "philosophy" on which they are built.

The task

For the purposes of these conceptual explanations, I will simplify the OP's task a bit: We need to make an input voltage that varies in the range of -5 V to 5 V…

^{simulate this circuit – Schematic created using CircuitLab}

… vary in the range of 0 V to 10 V.

^{simulate this circuit}

Basic idea

But is not it best to reveal the most general idea first, and then discuss its specific implementations? Because ideas are something eternal that does not depend on specific implementations…

Obviously, the idea here is to add a 5 V offset voltage Voff to the input voltage Vin (Vout = Vin + Voff). Then, when the input voltage varies between -5 V and 5 V, the output voltage will vary between 0 V and 10 V.

Implementation

Simply put, we need a summer to sum the two voltages. What is the simplest way to sum voltages?

Series summer

Apparently, in series according to the KVL is the best way. To do this, we connect the two voltage sources in series, and the whole circuit to the load.

^{simulate this circuit}

But what is the summer here? It is what remains after removing the input sources and the load, i.e. a piece of wire ("nothing").

^{simulate this circuit}

However, there is a problem - the offset source is floating (ungrounded). If we simply ground it (e.g., through a 1 Ω ammeter SHORT), a short circuit occurs and a high 5 A current flows through Vin.

^{simulate this circuit}

The short circuit disappears if we swap the two sources…

^{simulate this circuit}

… but the "floating" input source does not suit us either; both sources must be grounded. However, then the load will be "floating"... and we do not want that either.

^{simulate this circuit}

How do we solve this last problem? We need a device (an amplifier with a gain of 1) with a “floating” (differential) input and a grounded (single-ended) output.

^{simulate this circuit}

This is the solution I proposed to @Tyassin in my comment.

Parallel summer

The solution above needs an op-amp with differential input. But imagine that we have only an amplifier with a single-ended (grounded) input (that was a century ago).

Trying to connect directly Vin and Vref in parallel: Cannot we sum the voltages of the two sources by simply connecting them in parallel? It does not because they come into conflict and a high 5 A current flows.

^{simulate this circuit}

Mitigating the conflict: So, we have to mitigate the conflict by connecting resistors R1 and R2 in series to voltage sources. What is the summer here? It is a little more "complicated" - the two resistors R1 and R2. This humble circuit deserves more attention; so let's examine it manually at three input voltage values:

Vin = -5 V

^{simulate this circuit}

Vin = 0 V

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Vin = 5 V

^{simulate this circuit}

Finally, let's sweep Vin. What we see is that this circuit both attenuates (two times) and sums the input voltages, i.e. it is a weighted-input summer. As you can see, Vout (in blue) is two times smaller than above (ignore the orange curve; it is only to "cheat" the simulator autoscaling).

Amplified by a non-inverting amplifier: Then let's amplify 2x the output voltage.

^{simulate this circuit}

This is the @Tyassin's solution.

Op-amp inverting summer: With the same success we can amplify the output voltage (current) with an inverting (transimpedance) amplifier.

^{simulate this circuit}

The only problem is that it is inverted.

This is the left part of the @Math Keeps Me Busy's circuit solution.