The crux of the question is
I'm talking about a universe in which wave functions have only real parts with no imaginary parts, a universe described by something different from the Schrödinger equation for the analog of non-relativistic quantum mechanics, and/or a universe described by something different from the Dirac equation for the analog of relativistic quantum mechanics.
These ideas are all, I believe, tied together, and at the root of the answer is simply the axiom that probability must be conserved. From this, along with a couple of other assumptions, you can show that wave functions cannot be purely real-valued. From this, the form of the Schrödinger equation becomes apparent.
Why the wave function must be complex
Let's say that at time $t=0$, a particle is in the state $|\psi(0)\rangle$ (where I am describing quantum states in bra-ket notation)$^{\dagger}$. There must be some operator $\hat{U}(t)$, known as the time-evolution operator, that shows how this state evolves in time, i.e. such that for any future state $|\psi(t)\rangle$, we can write
$$|\psi(t)\rangle=\hat{U}(t)|\psi(0)\rangle$$
Now say that the particle is in a state $|\psi_1\rangle$, and we wish to find the probability that it is in $|\psi_2\rangle$, which we denote by $\langle\psi_2|\psi_1\rangle$. Naturally, if the two states are identical, this probability must be 1: There is a 100% chance of finding the particle in the state it is in. Therefore, we require the following to be true:
$$\langle\psi(t)|\psi(t)\rangle=1=\langle\psi(0)|\psi(0)\rangle$$
But we can invoke the time-evolution operator to rewrite $|\psi(t)\rangle$ and see that
$$\langle\psi(0)|\hat{U}^{\dagger}(t)\hat{U}(t)|\psi(0)\rangle=\langle\psi(0)|\psi(0)\rangle$$
where $\hat{U}^{\dagger}(t)$ is known as the adjoint of the operator. For the above equation to be true, we need $\hat{U}^{\dagger}(t)\hat{U}(t)=1$, which is the definition of a unitary operator. If this holds, probability is conserved.
This is where complex numbers enter the stage. We can show that any unitary operator can be written in the form of a complex exponential; because $\hat{U}(t)$ is unitary, it obeys that line of reasoning, and as such must be complex. In quantum mechanics, it happens to take the form
$$\hat{U}(t)=e^{-i\hat{H}t/\hbar}$$
with $\hat{H}$ the operator known as the Hamiltonian and $\hbar$ is the reduced Planck constant. We see immediately that, in general, $|\psi(t)\rangle$ must be complex.$^{\ddagger}$
For further reading, see About the complex nature of the wave function? and QM without complex numbers on Physics Stack Exchange. Some of those answers use empirical arguments, but pcr's answer makes the same argument as mine and remains purely theoretical, and is by extension still applicable to your universe.
The Schrödinger equation from $\hat{U}(t)$
From the time-evolution operator, we can quickly derive a form of the Schrödinger equation by looking at an infinitesimal time translation
$$\hat{U}(dt)=1-\frac{i}{\hbar}\hat{H}dt$$
At a time $t+dt$, we can find the state of the system from $\hat{U}(t+dt)$, which you can convince yourself is just $\hat{U}(dt)\hat{U}(t)$:
$$\hat{U}(t+dt)=\left(1-\frac{i}{\hbar}\hat{H}dt\right)\hat{U}(t)$$
Rearranging,
$$\hat{U}(t+dt)-\hat{U}(t)=\left(-\frac{i}{\hbar}\hat{H}\right)\hat{U}(t)$$
If we divide both sides by $dt$, we see that the left just gives us the expression for the time derivative of $\hat{U}(t)$. We can then rewrite this as
$$i\hbar\frac{d}{dt}\hat{U}=\hat{H}\hat{U}(t)$$
Applying both sides to the initial state $|\psi(0)\rangle$ gives us
$$i\hbar\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$$
which is the Schrödinger equation. This is a quick-and-dirty derivation (source: Townsend, A Modern Approach to Quantum Mechanics, second edition, chapter 4).
The Dirac equation
The Dirac equation is much more complicated. It breaks the wave function into four separate components, and in reality, it's actually four separate coupled first-order linear partial differential equations. I'm not as familiar with the Dirac equation as I am with the Schrödinger equation, so I won't try to do it justice, but I will say that given that it can be thought as arising from taking the square root, so to speak, of the operator
$$\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$
we can see that perhaps a factor of $i$ must creep in somewhere to account for that minus sign.
$^{\dagger}$The wave function $\psi$ can be found from a quantum state by taking the appropriate inner product. For example, if we want the position space representation of the wave function, we define it by the inner product $\psi(x)\equiv\langle x|\psi\rangle$; if we want the momentum space representation, we use the inner product $\psi(p)\equiv\langle p|\psi\rangle$. While I've technically focused this answer on quantum states, it's simple to show that the logic by extension also holds for wave functions.
$^{\ddagger}$In the case where $\hat{H}=0$, we have $\hat{U}(t)=1$, and so if $|\psi(0)\rangle$ is real, then so is $|\psi(t)\rangle$. On the other hand, this is a trivial case that holds only under one particular (and extremely odd) set of circumstances, and in reality, no particle is truly subject to a vanishing Hamiltonian.
